My Favorite One-Liners: Part 5

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

Every once in a while, students encounter a step that seems far from obvious. To give one example, to evaluate the series

\displaystyle \sum_{n=1}^{100} \frac{1}{n^2+n},

the natural first step is to rewrite this as

\displaystyle \sum_{n=1}^{100} \left(\frac{1}{n} - \frac{1}{n+1} \right)

and then use the principle of telescoping series. However, students may wonder how they were ever supposed to think of the first step for themselves.

Students often give skeptical, quizzical, and/or frustrated looks about this non-intuitive next step… they’re thinking, “How would I ever have thought to do that on my own?” To allay these concerns, I explain that this step 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.

Sadly, there aren’t any videos of Greek philosophers teaching, so I’ll have to settle for this:

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 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 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}

green lineThe 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.