# Non-geometric infinite series (Part 12)

I conclude this series of posts with thoughts about infinite series which use reciprocals of positive integers. I offer this post for the enrichment of talented Precalculus students who have exhibited mastery of geometric series.

Geometric. As we’ve discussed at length, the series

$1 + \displaystyle \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots$

converges and is in fact equal to $2$.

Harmonic. Including the reciprocals of all positive integers is called a harmonic series:

$1 + \displaystyle \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \dots$

As shown in the link to the MathWorld website, this series actually diverges, even though the terms get smaller and smaller.

So we’ve made an observation: if too many reciprocals are included, the series diverges. But if we take enough of them away, then we can still end up with a series that is infinite but converges.

Squares. Let’s now consider the reciprocals of perfect squares:

$1 + \displaystyle \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \dots$

Clearly, we’ve taken away a lot of the terms of the harmonic series? Have we taken enough away so that the series converges? It turns out that the answer is yes. And the answer is precisely what you’d think it should be (not): $\pi^2/6$. This is just another way that the circumference of a circle has an odd way of appearing in the most unexpected of places.

The proof that this series equals $\pi^2/6$ requires the clever use of Parseval’s theorem from Fourier analysis.

Fourth Powers. Let’s now turn to the reciprocals of fourth powers:

$1 + \displaystyle \frac{1}{16} + \frac{1}{81} + \frac{1}{256} + \frac{1}{625} + \dots$

By the Direct Comparison Test and the series for reciprocals of squares, this series converges. Using Parseval’s theorem, it can be shown that the answer is $\pi^4/90$.

Cubes. Now let’s investigate the reciprocals of cubes:

$1 + \displaystyle \frac{1}{8} + \frac{1}{27} + \frac{1}{64} + \frac{1}{125} + \dots$

Again by the Direct Comparison Test and the series for reciprocals for squares, this series must converge. This sum is called Apéry’s constant.  However, and amazingly, no one knows what the answer is. Of course, a computer can be programmed to evaluate this series to as many decimal places as desired. According to Wikipedia, this sum was evaluated to over 100 billion decimal places in 2010. However, to the best of my knowledge, no one has figured out if there’s a simple way of writing the answer, like $\pi^2/6$ or $\pi^4/90$.

So if you figure out a simple way to evaluate Apéry’s constant, feel free to call me collect.

The previous four series are example of Riemann’s zeta function, which is of central importance in number theory and is the focus of the celebrated Riemann Hypothesis, for which a solution is worth a cool \$1 million.

Primes. Now let’s consider the reciprocals of primes:

$\displaystyle \frac{1}{2} + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \frac{1}{11} + \dots$

As noted above, the harmonic series diverges, but if we remove enough terms from the harmonic series, then it’s possible to make an infinite series that converges. So the central question is, did we remove enough fractions (by taking away all of the composite denominators) so that the series converges?

Surprisingly, the answer is no: the sum of the reciprocals of the primes actually diverges. The proof actually requires a graduate-level class in analytic number theory.

By no means would I expect high school students to master all of the above facts. As noted above, the subject of this post is mostly for the enrichment of high school students who have mastered infinite geometric series.

That said, students who know the following facts from Precalculus will be well-served when they reach calculus and other university-level mathematics courses.

• They should either know the formula for an infinite geometric series or else be able to quickly derive it.
• They should know that not every infinite geometric series is geometric.
• They should know that not every infinite series converges.
• They should be familiar with the meaning of the terms converge and diverge.

# Formula for an infinite geometric series (Part 11)

Many math majors don’t have immediate recall of the formula for an infinite geometric series. They often can remember that there is a formula, but they can’t recollect the details. While it’s I think it’s OK that they don’t have the formula memorized, I think is a real shame that they’re also unaware of where the formula comes from and hence are unable to rederive the formula if they’ve forgotten it.

In this post, I’d like to give some thoughts about why the formula for an infinite geometric series is important for other areas of mathematics besides Precalculus. (There may be others, but here’s what I can think of in one sitting.)

1. An infinite geometric series is actually a special case of a Taylor series. (See https://meangreenmath.com/2013/07/05/reminding-students-about-taylor-series-part-5/ for details.) Therefore, it would be wonderful if students learning Taylor series in Calculus II could be able to relate the new topic (Taylor series) to their previous knowledge (infinite geometric series) which they had already seen in Precalculus.

2. An infinite geometric series is also a special case of the binomial series $(1+x)^n$, when $n$ does not have to be a positive integer and hence Pascal’s triangle cannot be used to find the expansion.

3. Infinite geometric series is a rare case when an infinite sum can be found exactly. In Calculus II, a whole battery of tests (e.g., the Root Test, the Ratio Test, the Limit Comparison Test) are introduced to determine whether a series converges or not. In other words, these tests only determine if an answer exists, without determining what the answer actually is.

Throughout the entire undergraduate curriculum, I’m aware of only four types of series that can actually be evaluated exactly.

• An infinite geometric series with $-1 < r < 1$
• The Taylor series of a real analytic function. (Of course, an infinite geometric series is a special case of a Taylor series.)
• A telescoping series. For example, using partial fractions and cancelling a bunch of terms, we find that

$\displaystyle \sum_{k=1}^\infty \frac{1}{k^2+k} = \displaystyle \sum_{k=1}^\infty \left( \frac{1}{k} - \frac{1}{k+1} \right)$

$\displaystyle \sum_{k=1}^\infty \frac{1}{k^2+k} = \displaystyle \left( 1 - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) \dots$

$\displaystyle \sum_{k=1}^\infty \frac{1}{k^2+k} = 1$

4. Infinite geometric series are essential for proving basic facts about decimal representations that we often take for granted.

5. Properties of an infinite geometric series are needed to find the mean and standard deviation of a geometric random variable, which is used to predict the number of independent trials needed before an event happens. This is used for analyzing the coupon collector’s problem, among other applications.

# Formula for an infinite geometric series (Part 10)

I conclude this series of posts by considering the formula for an infinite geometric series. Somewhat surprisingly (to students), the formula for an infinite geometric series is actually easier to remember than the formula for a finite geometric series.

One way of deriving the formula parallels the derivation for a finite geometric series. If $a_1, a_2, a_3, \dots$ are the first terms of an infinite geometric sequence, let

$S = a_1 + a_2 + a_3 + \dots$

Recalling the formula for an geometric sequence, we know that

$a_2 = a_1 r$

$a_3 = a_1 r^2$

$\vdots$

Substituting, we find

$S = a_1 + a_1 r+ a_1 r^2 \dots$

Once again, we multiply both sides by $-r$.

$-rS = -a_1r - a_1 r^2- a_1 r^3 \dots$

Next, we add the two equations. Notice that almost everything cancels on the right-hand side… except for the leading term $a_1$.  (Unlike yesterday’s post, there is no “last” term that remains since the series is infinite.) Therefore,

$S - rS = a_1$

$S(1-r) = a_1$

$S = \displaystyle \frac{a_1}{1-r}$

A quick pedagogical note: I find that this derivation “sells” best to students when I multiply by $-r$ and add, as opposed to multiplying by $r$ and subtracting.

The above derivation is helpful for remembering the formula but glosses over an extremely important detail: not every infinite geometric series converges. For example, if $a_1 = 1$ and $r = 2$, then the infinite geometric series becomes

$1 + 2 + 4 + 8 + 16 + \dots$,

which clearly does not have a finite answer. We say that this series diverges. In other words, trying to evaluate this sum makes as much sense as trying to divide a number by zero: there is no answer.

That said, it can be shown that, as long as $-1 < r < 1$, then the above geometric series converges, so that

$a_1 + a_1 r + a_1 r^2 + \dots = \displaystyle \frac{a_1}{1-r}$

The formal proof requires the use of the formula for a finite geometric series:

$a_1 + a_1 r + a_1 r^2 + \dots + a_1 r^{n-1} = \displaystyle \frac{a_1(1-r^n)}{1-r}$

We then take the limit as $n \to \infty$:

$\displaystyle \lim_{n \to \infty} a_1 + a_1 r + a_1 r^2 + \dots + a_1 r^{n-1} = \displaystyle \lim_{n \to \infty} \frac{a_1(1-r^n)}{1-r}$

$a_1 + a_1 r + a_1 r^2 + \dots = \displaystyle \lim_{n \to \infty} \frac{a_1(1-r^n)}{1-r}$

On the right-hand side, the only piece that contains an $n$ is the term $r^n$. If $-1 < r < 1$, then $r^n \to 0$ as $n \to \infty$. (This limit fails, however, if $r \ge 1$ or $r \le -1$.) Therefore,

$a_1 + a_1 r + a_1 r^2 + \dots = \displaystyle \lim_{n \to \infty} \frac{a_1(1-0)}{1-r} = \displaystyle \frac{a_1}{1-r}$

# Formula for an infinite geometric series (Part 9)

I continue this series of posts by considering the formula for an infinite geometric series. Somewhat surprisingly (to students), the formula for an infinite geometric series is actually easier to remember than the formula for a finite geometric series.

One way of deriving the formula parallels yesterday’s post. If $a_1, a_2, a_3, \dots$ are the first terms of an infinite geometric sequence, let

$S = a_1 + a_2 + a_3 + \dots$

Recalling the formula for an geometric sequence, we know that

$a_2 = a_1 r$

$a_3 = a_1 r^2$

$\vdots$

Substituting, we find

$S = a_1 + a_1 r+ a_1 r^2 \dots$

For example, if $a_1 = \displaystyle \frac{1}{2}$ and $r = \displaystyle \frac{1}{2}$, we have

$S = \displaystyle \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots$

This is perhaps the world’s most famous infinite series, as this is the subject of Zeno’s paradox. When I teach infinite series in class, I often engage the students by reminding students about Zeno’s paradox and then show them this clip from the 1994 movie I.Q.

This clip is almost always a big hit with my students.

Even after showing this clip, some students resist the idea that an infinite series can have a finite answer. For such students, I use a physical demonstration: I walk half-way across the classroom, then a quarter, and so on… until I walk head-first into a walk at full walking speed. The resulting loud thud usually confirms for students that an infinite sum can indeed have a finite answer.

P.S. PhD Comics recently had a cartoon concerning Zeno’s paradox. Source: http://www.phdcomics.com/comics/archive.php?comicid=1610

Here’s another one. Source: http://www.xkcd.com/994/

# Formula for a finite geometric series (Part 8)

As I’ve said before, I’m not particularly a fan of memorizing formulas. Apparently, most college students aren’t fans either, because they often don’t have immediate recall of certain formulas from high school when they’re needed in the collegiate curriculum.

While I’m not a fan of making students memorize formulas, I am a fan of teaching students how to derive formulas. Speaking for myself, if I ever need to use a formula that I know exists but have long since forgotten, the ability to derive the formula allows me to get it again.

Which leads me to today’s post: the derivation of the formulas for the sum of a finite geometric series. This topic is commonly taught in Precalculus but, in my experience, is often forgotten by students years later when needed in later classes.

Like its counterpart for arithmetic series, the formula for a finite geometric series can be derived using the patented Bag of Tricks. Socrates gave the Bag of Tricks to Plato, Plato gave it to Aristotle, it passed down the generations, my teacher taught the Bag of Tricks to me, and I teach it to my students.

If $a_1, \dots, a_n$ are the first $n$ terms of an geometric sequence, let

$S = a_1 + a_2 + a_3 + \dots + a_{n-1} + a_n$

Recalling the formula for an geometric sequence, we know that

$a_2 = a_1 r$

$a_3 = a_1 r^2$

$\vdots$

$a_{n-1} = a_1 r^{n-2}$

$a_n = a_1 r^{n-1}$

Substituting, we find

$S = a_1 + a_1 r+ a_1 r^2 \dots + a_1 r^{n-2} + a_1 r^{n-1}$

At this point, we use something different from the patented Bag of Tricks: we multiply both sides by $-r$.

$-rS = -a_1r - a_1 r^2- a_1 r^3 \dots - a_1 r^{n-1} - a_1 r^n$

Next, we add the two equations. Notice that almost everything cancels on the right-hand side. The $a_1 r$ cancel, the $a_1 r^2$ cancel, yada yada yada, and the $a_1 r^{n-1}$ cancel. The only terms that remain are $a_1$ and $-a_1 r^n$. So

$S - rS = a_1 - a_1 r^n$

$S(1-r) = a_1 (1- r^n)$

$S = \displaystyle \frac{a_1 ( 1-r^n) }{1-r}$

A quick pedagogical note: I find that this derivation “sells” best to students when I multiply by $-r$ and add, as opposed to multiplying by $r$ and subtracting.

This formula is also a straightforward consequence of the factorization formula

$x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + \dots +x y^{n-2} + y^{n-1})$

Just let $x=1$ and $y=r$, and then multiply both sides by the first term $a_1$.

However, in my experience, most students don’t have instant recall of this formula either. They can certainly remember the formula for the difference of two squares (which is a special case of the above formula), but they often can’t remember that the difference of two cubes has a formula. (And, while I’m on the topic, they also can’t remember that the sum of two cubes can always be factored.)

Pedagogically, the most common mistake that I see students make when using this formula is using the wrong exponent on the right-hand side. For example, suppose the problem is to simplify

$48 +24+ 12 + 6 + 3 + 1.5 + 0.75 + 0.375$

Here’s the common mistake: student solve for $n$ using the formula for a geometric sequence. They solve for the unknown exponent (often using logarithms) and find that $0.375 = 48 (0.5)^7$. They conclude that $n=7$, and then plug into for formula for a geometric series:

$S = \displaystyle \frac{48 ( 1-(0.5)^7) }{1-0.5} = 95.25$ (incorrect)

This answer is clearly wrong, since the sum of the original series must have a $5$ in the thousandths place. The answer $95.25$ is the correct answer to the wrong question — that’s the sum of the first seven terms of the sequence (stopping at $0.75$), but the original series has eight terms. Using the formula correctly, we find

$S = \displaystyle \frac{48 ( 1-(0.5)^8) }{1-0.5} = 95.625$ (correct)

Not surprisingly, the difference between the incorrect and correct answers is $0.375$, the eighth term.

To help students avoid this mistake, I re-emphasize that the number $n$ stands for the number of terms in the series. In particular, it does not mean the exponent needed to give the last term in the series. That exponent, of course, is $n-1$, not $n$.

# Formula for an arithmetic series (Part 7)

As we’ve discussed, the formula for an arithmetic series is

$S_n = \displaystyle \frac{n}{2} (2a_1 + [n-1] d) = \displaystyle \frac{n}{2} (a_1 + a_n)$,

where $n$ is the number of terms, $a_1$ is the first term, $d$ is the common difference, and $a_n$ is the last term. This formula may be more formally expressed as

$S = \displaystyle \sum_{k=1}^n a_k = \displaystyle \frac{n}{2} (2a_1 + [n-1] d) = \displaystyle \frac{n}{2} (a_1 + a_n)$

For homework and on tests, students are asked to directly plug into this formula and to apply this problem with word problems, like finding the total number of seats in an auditorium with 50 rows, where there are 12 seats in the front row and each row has two more seats than the row in front of it.

In my opinion, the ability to solve questions like the one below is the acid test for determining whether a student — who I assume can solve routine word problems like the one above — really understands series or is just familiar with series. In other words, if a student can solve routine word problems but is unable to handle a problem like the one below, then there’s still room for that student’s knowledge of series to deepen.

Calculate $\displaystyle \sum_{k=11}^{60} (5k - 2)$

There are two reasonable approaches for solving this problem.

Solution #1. Notice that $5k - 2 = 5(k-1) + 5 - 2 = 3 + 5(k-1)$. So this is really an arithmetic series whose first term is $3$ and whose common difference is $5$. Therefore,

$S = \displaystyle \sum_{k=1}^{60} a_k = \displaystyle \frac{60}{2} (2[3] + [60-1] 5)=9030$

However, I’m supposed to start the series on $k=11$, not $k=1$. That means that I need to subtract off the first ten terms of the above series. Now

$S = \displaystyle \sum_{k=1}^{10} a_k = \displaystyle \frac{10}{2} (2[3] + [10-1] 5)= 255$

Finally,

$\displaystyle \sum_{k=11}^{60} a_k = \displaystyle \sum_{k=1}^{60} a_k - \displaystyle \sum_{k=1}^{10} a_k = 9030 - 255 = 8775$

Solution #2. Writing out the terms, we see that

$\displaystyle \sum_{k=11}^{60} (5k - 2) = (5[11]-2) + (5[12]-2) + \dots + (5[60]-2)$

or

$\displaystyle \sum_{k=11}^{60} (5k - 2) = 53+58 + \dots +298$

The right-hand side is an arithmetic series whose “first” term is $53$ and whose last ($50$th) term is $298$. Therefore,

$\displaystyle \sum_{k=11}^{60} (5k - 2) = \frac{50}{2} (53+298) = 8775$

Of the two solutions, I suppose I have a mild preference for the first, as the second solution won’t work for something like $\displaystyle \sum_{k=11}^{60} k^2$. However, both solution demonstrate that the student is actually thinking about the meaning of the series instead of just plugging numbers in a formula, and so I’d be happy with either one in a Precalculus class.

# Formula for an arithmetic series (Part 6)

In the previous posts of this series, I described two methods of deriving the formula

$\displaystyle \sum_{k=1}^n k = \frac{n(n+1)}{2}$

The first method concerned reversing the terms of the sum (or, almost equivalently, taking the terms in pairs). The second method used mathematical induction.

Mathematical induction can be applied to arithmetic series as well as other series. However, the catch is that you have to know the answer before proving that the answer actually is correct. By contrast, the first method did not require us to know the answer in advance — it just fell out of the calculation — but it cannot be applied to series that are not arithmetic.

Here’s a third method using the principle of telescoping series. This method has the strengths of the previous two methods: it does not require us to know the answer in advance, and it can also be applied to some other series which are not arithmetic.

To begin, consider the sum

$\displaystyle \sum_{k=1}^n [k^2 - (k-1)^2]$

At this early point, students often object, “Where did that come from?” I’ve said it before but I’ll say it again: I tell them my usual tongue-in-cheek story that this idea comes from the patented Bag of Tricks. Socrates gave the Bag of Tricks to Plato, Plato gave it to Aristotle, it passed down the generations, my teacher taught the Bag of Tricks to me, and I teach it to my students.

In any event, I will evaluate this sum in two different ways.

Step 1. Just write out the terms of the series, starting from $k=1$ and ending with $k =n$.

$\displaystyle \sum_{k=1}^n [k^2 - (k-1)^2] = [1^2 - 0^2] + [2^2 - 1^2] + [3^2 - 2^2] + \dots + [n^2 - (n-1)^2]$

Notice that, on the right-hand side, the $1^2$ terms cancel, the $2^2$ terms cancel, and so on. In fact, almost everything cancels. The only two terms that aren’t cancelled are the $0^2$ and $n^2$ terms. Therefore,

$\displaystyle \sum_{k=1}^n [k^2 - (k-1)^2] = n^2 - 0^2 = n^2$

Step 2. Next, we’ll rewrite the original sum by expanding out the terms inside of the sum:

$\displaystyle \sum_{k=1}^n [k^2 - (k-1)^2] = \displaystyle \sum_{k=1}^n [k^2 - (k^2 -2k + 1)]$

$\displaystyle \sum_{k=1}^n [k^2 - (k-1)^2] = \displaystyle \sum_{k=1}^n [2k-1]$

$\displaystyle \sum_{k=1}^n [k^2 - (k-1)^2] = \displaystyle \sum_{k=1}^n 2k - \displaystyle \sum_{k=1}^n 1$

$\displaystyle \sum_{k=1}^n [k^2 - (k-1)^2] = 2\displaystyle \sum_{k=1}^n k - \displaystyle \sum_{k=1}^n 1$

Step 3. Of course, these different looking answers from Steps 1 and 2 have to be the same, so let’s set them equal to each other:

$2\displaystyle \sum_{k=1}^n k - \displaystyle \sum_{k=1}^n 1 = n^2$

There is one unknown in this equation, $\displaystyle \sum_{k=1}^n k$. The second sum is just the constant $1$ added to itself $n$ times, and so $\displaystyle \sum_{k=1}^n 1 = n$. Therefore, we solve for the unknown:

$2 \left(\displaystyle \sum_{k=1}^n k \right) - n = n^2$

$2 \left(\displaystyle \sum_{k=1}^n k \right) = n^2 + n$

$\displaystyle \sum_{k=1}^n k = \displaystyle \frac{n^2 + n}{2}$

The beauty of this approach is that this approach can be continued. For example, to obtain $\displaystyle \sum_{k=1}^n k^2$, we begin with

$\displaystyle \sum_{k=1}^n [k^3 - (k-1)^3]$

Step 1. By telescoping series,

$\displaystyle \sum_{k=1}^n [k^3 - (k-1)^3] = n^3 - 0^3 = n^3$

Step 2. Using the binomial theorem,

$\displaystyle \sum_{k=1}^n [k^3 - (k-1)^3] = \displaystyle \sum_{k=1}^n [k^3 - (k^3 -3k^2+3k- 1)]$

$\displaystyle \sum_{k=1}^n [k^2 - (k-1)^2] = 3\displaystyle \sum_{k=1}^n k^2 - 3\displaystyle \sum_{k=1}^n k + \displaystyle \sum_{k=1}^n 1$

$\displaystyle \sum_{k=1}^n [k^2 - (k-1)^2] = 3\displaystyle \sum_{k=1}^n k^2 - 3\left( \frac{n(n+1)}{2} \right) + n$

Step 3. Setting these two expressions equal to each other,

$3\displaystyle \sum_{k=1}^n k^2 - 3\left( \frac{n(n+1)}{2} \right) + n= n^3$

And we eventually conclude that:

$\displaystyle \sum_{k=1}^n k^2 = \displaystyle \frac{n(n+1)(2n+1)}{6}$

And then this could be continued to obtain closed-form expressions for higher exponents of $k$.

# Formula for an arithmetic series (Part 5)

In Precalculus, Discrete Mathematics or Real Analysis, an arithmetic series is often used as a student’s first example of a proof by mathematical induction. Recall, from Wikipedia:

Mathematical induction is a method of mathematical proof typically used to establish a given statement for all natural numbers. It is done in two steps. The first step, known as the base case, is to prove the given statement for the first natural number. The second step, known as the inductive step, is to prove that the given statement for any one natural number implies the given statement for the next natural number. From these two steps, mathematical induction is the rule from which we infer that the given statement is established for all natural numbers.

The simplest and most common form of mathematical induction infers that a statement involving a natural number n holds for all values of n. The proof consists of two steps:

1. The basis (base case): prove that the statement holds for the first natural number $n$. Usually, $n=0$ or $n=1$.
2. The inductive step: prove that, if the statement holds for some natural number $n$, then the statement holds for $n+1$.

The hypothesis in the inductive step that the statement holds for some $n$ is called the induction hypothesis (or inductive hypothesis). To perform the inductive step, one assumes the induction hypothesis and then uses this assumption to prove the statement for $n+1$.

As an inference rule, mathematical induction can be justified as follows. Having proven the base case and the inductive step, then any value can be obtained by performing the inductive step repeatedly. It may be helpful to think of the domino effect. Consider a half line of dominoes each standing on end, and extending infinitely to the right. Suppose that:

1. The first domino falls right.
2. If a (fixed but arbitrary) domino falls right, then its next neighbor also falls right.

With these assumptions one can conclude (using mathematical induction) that all of the dominoes will fall right.

Mathematical induction… works because $n$ is used to represent an arbitrary natural number. Then, using the inductive hypothesis, i.e. that $P(n)$ is true, show $P(k+1)$ is also true. This allows us to “carry” the fact that $P(0)$ is true to the fact that $P(1)$ is also true, and carry $P(1)$ to $P(2)$, etc., thus proving $P(n)$ holds for every natural number $n$.

When students first encounter mathematical induction (in either Precalculus, Discrete Mathematics, or Real Analysis), the theorems that students are asked to prove usually fall into four categories:

1. Calculating a series (examples below).
2. Statements concerning divisibility (for example, proving that $4$ is always a factor of $5^n-1$).
3. Finding a closed-form expression for a recursively defined sequence (for example, if $a_1 = 4$ and $a_n = 3a_{n-1}$ if $n \ge 2$, proving that $a_n = 4 \times 3^{n-1}$0
4. Statements concerning inequality (for example, proving that $n! > 4^n$ if $n \ge 9$0

Here’s a common first example of mathematical induction applied to an arithmetic series. Notice that the statement of the theorem matches the form $\displaystyle \frac{n}{2}(a_1 + a_n)$ seen earlier in this series (pardon the pun) of posts.

Theorem. $1 + 2 + \dots + (n-1) + n = \displaystyle \frac{n(n+1)}{2}$

Proof. Induction on $n$.

$n = 1$: The left-hand is simply $1$, while the right-hand side is $\displaystyle \frac{(1)(2)}{2}$, which is also equal to $1$. So the base case works.

$n$: Assume that the statement holds true for the integer $n$.

$n+1$. If I replace $n$ by $n+1$ in the statement of the theorem, then the right-hand side becomes

$\displaystyle \frac{(n+1)[(n+1)+1]}{2} = \displaystyle \frac{(n+1)(n+2)}{2}$

I find it helpful to describe this to students as my target. In other words, as I manipulate the left-hand side, my ultimate goal is to end up with this target. Once I have done that, then I have completed the proof.

If I replace $n$ by $n+1$ in the statement of the theorem, then the left-hand side will now end on $n+1$ instead of $n$:

$1 + 2 + \dots + (n-1) + n + (n+1)$

Notice that we’ve seen almost all of this before, except for the extra term $n+1$. So we will substitute using the induction hypothesis, carrying the extra $n+1$ along for the ride.

$1 + 2 + \dots + (n-1) + n + (n+1) = \displaystyle \frac{n(n+1)}{2} + (n+1)$

Now our task is, by hook or by crook, using whatever algebraic tricks we can think of to convert this last expression into the target. Most students are completely comfortable doing this, although they typically multiply out the term $n(n+1)$ unnecessarily. Indeed, many early proofs by induction are simplified by factoring out terms whenever possible — in the example below, $(n+1)$ is factored on the last step — as opposed to multiplying them out. In my experience, proofs by induction often serve as a stringent test of students’ algebra skills as opposed to their skills in abstract reasoning.

In any event, here’s the end of the proof:

$1 + 2 + \dots + (n-1) + n + (n+1) = \displaystyle \frac{n(n+1)}{2} + (n+1)$

$1 + 2 + \dots + (n-1) + n + (n+1) = \displaystyle \frac{n(n+1) + 2(n+1)}{2}$

$1 + 2 + \dots + (n-1) + n + (n+1) = \displaystyle \frac{(n+1)(n + 2)}{2}$

Mathematical induction can be used to verify formulas for series which are not arithmetic, like

$1^2 + 2^2 + \dots + (n-1)^2 + n^2 = \displaystyle \frac{n(n+1)(2n+1)}{6}$

$1^3 + 2^3 + \dots + (n-1)^3 + n^3 = \displaystyle \frac{n^2(n+1)^2}{4}$

However, the downside of a proof by induction lies in the word verify, as it’s necessary to actually know what’s going to work before proceeding with the proof.

In the next post, I’ll describe a method of obtaining these series that does not require mathematical induction.

# Formula for an arithmetic series (Part 4)

As I’ve said before, I’m not particularly a fan of memorizing formulas. Apparently, most college students aren’t fans either, because they often don’t have immediate recall of certain formulas from high school when they’re needed in the collegiate curriculum.

While I’m not a fan of making students memorize formulas, I am a fan of teaching students how to derive formulas. Speaking for myself, if I ever need to use a formula that I know exists but have long since forgotten, the ability to derive the formula allows me to get it again.

Which leads me to today’s post: the derivation of the formulas for the sum of an arithmetic series. This topic is commonly taught in Precalculus but, in my experience, is often forgotten by students years later when needed in later classes.

To get the idea across, consider the arithmetic series

$S = 16 + 19 + 22 + 25 + 28 + 31 + 34 + 37 + 40 + 43$

Now write the sum in reverse order. This doesn’t change the value of the sum, and so:

$S = 43 + 40 + 37 +34+ 31 + 28 + 25 + 22 + 19 + 16$

Now add these two lines vertically. Notice that $16 + 43 = 59$, $19 + 40 = 59$, and in fact each pair of numbers adds to $59$. So

$2S = 59 + 59 + 59 + 59 + 59 + 59 + 59 + 59 + 59 + 59$

$2S = 59 \times 10 = 590$

$S = 295$

Naturally, this can be directly confirmed with a calculator by just adding the 10 numbers.

When I show this to my students, they often complain that there’s no way on earth that they would have thought of that for themselves. They wouldn’t have thought to set the sum equal to $S$, and they certainly would not have thought to reverse the terms in the sum. To comfort them, I tell them my usual tongue-in-cheek story that this idea comes from the patented Bag of Tricks. Socrates gave the Bag of Tricks to Plato, Plato gave it to Aristotle, it passed down the generations, my teacher taught the Bag of Tricks to me, and I teach it to my students.

The derivation of the general formula proceeds using the same idea. If $a_1, \dots, a_n$ are the first $n$ terms of an arithmetic sequence, let

$S = a_1 + a_2 + \dots + a_{n-1} + a_n$

Recalling the formula for an arithmetic sequence, we know that

$a_2 = a_1 + d$

$\vdots$

$a_{n-1} = a_1 + (n-2)d$

$a_n = a_1 + (n-1)d$

Substituting, we find

$S = a_1 + [a_1 + d] + \dots + [a_1 + (n-2)d] + [a_1 + (n-1)d]$

As above, we now return the order…

$S = [a_1 + (n-1)d] + [a_1 + (n-2)d] + \dots + [a_1 + d] + a_1$

… and add the two equations:

$2S = [2a_1 + (n-1)d] + [2a_1 + d+(n-2)d] + \dots + [2a_1 +(n-2)d+ d] + [2a_1+(n-1)d]$

$2S = [2a_1 + (n-1)d] + [2a_1 + (n-1)d] + \dots + [2a_1 +(n-1)d] + [2a_1+(n-1)d]$

$2S = n[2a_1 + (n-1)d]$

$S = \displaystyle \frac{n}{2} [2a_1 + (n-1)d]$

We also note that the formula may be rewritten as

$S = \displaystyle \frac{n}{2} [a_1 + \{a_1 + (n-1)d\} ]$

or

$S = \displaystyle \frac{n}{2} [a_1 + a_n]$

This latter form isn’t too difficult to state as a sentence: the sum of a series with $n$  is the average of the first and last terms, multiplied by the number of terms.

Indeed, I have seen textbooks offer proofs of this formula by using the same logic that young Gauss used to find the sum $1 + 2 + \dots + 99 + 100$. The “proof” goes like this: Take the terms in pairs. The first term plus the last term is $a_1 + a_n$. The second term plus the second-to-last term is $a_2 + a_{n-1} = a_1 + d + a_n - d = a_1 + a_n$. And so on. So each pair adds to $a_1 + a_n$. Since there are $n$ terms, there are $n/2$ pairs, and so we derive the above formula for $S$.

You’ll notice I put “proof” in quotation marks. There’s a slight catch with the above logic: it only works if $n$ is an even number. If $n$ is odd, the result is still correct, but the logic to get the result is slightly different. That’s why I don’t particularly recommend using the above paragraph to prove this formula for students, even though it fits nicely with the almost unforgettable Gauss story.

That said, for talented students looking for a challenge, I would recommend showing this idea, then point out the flaw in the argument, and then ask the students to come up with an alternate proof for handling odd values of $n$.