Thoughts on Infinity (Part 3e)

In recent posts, we’ve seen the curious phenomenon that the commutative and associative laws do not apply to a conditionally convergent series or infinite product. Here’s another classic example of this fact that’s attributed to Cauchy.

We’ve already seen in this series (pardon the pun) that

$\displaystyle 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - ... = \ln 2$.

Let’s now see what happens if I rearrange the terms of this conditionally convergent series. Let

$T = \displaystyle 1 + \frac{1}{3} - \frac{1}{2} + \frac{1}{5} + \frac{1}{7} - \frac{1}{4} + \frac{1}{9} + \frac{11} - \frac{1}{6} \dots$,

where two positive numbers alternate with a single negative term. By all rights, this shouldn’t affect anything… right?

Let $s_n$ be the $n$th partial sum of this series, so that $s_{3n}$ contains $2n$ positive terms with odd denominators and $n$ negative terms with even denominators:

$s_{3n} = \displaystyle \sum_{k=1}^{2n} \frac{1}{2n-1} - \sum_{k=1}^n \frac{1}{2n}$.

Let me now add and subtract the “missing” even terms in the first sum:

$s_{3n} = \displaystyle \sum_{k=1}^{2n} \frac{1}{2n-1} + \sum_{k=1}^{2n} \frac{1}{2n} - \sum_{k=1}^{2n} \frac{1}{2n} - \sum_{k=1}^n \frac{1}{2n}$

$s_{3n} = \displaystyle \sum_{k=1}^{4n} \frac{1}{n} - \sum_{k=1}^{2n} \frac{1}{2n} - \sum_{k=1}^n \frac{1}{2n}$

$s_{3n} = \displaystyle \sum_{k=1}^{4n} \frac{1}{n} - \frac{1}{2} \sum_{k=1}^{2n} \frac{1}{n} - \frac{1}{2} \sum_{k=1}^n \frac{1}{n}$.

For reasons that will become apparent, I’ll now rewrite this as

$s_{3n} = \displaystyle \int_1^{4n} \frac{dx}{x} + \left( \sum_{k=1}^{4n} \frac{1}{k} - \displaystyle \int_1^{4n} \frac{dx}{x} \right)$

$- \displaystyle \frac{1}{2} \int_1^{2n} \frac{dx}{x} - \frac{1}{2} \left( \sum_{k=1}^{2n} \frac{1}{k} - \int_1^{2n} \frac{dx}{x} \right)$

$- \displaystyle \frac{1}{2} \int_1^{n} \frac{dx}{x} - \frac{1}{2} \left( \sum_{k=1}^{n} \frac{1}{k} - \int_1^{n} \frac{dx}{x} \right)$,

or

$s_{3n} = \ln(4n) - \ln 1 + \displaystyle \left( \sum_{k=1}^{4n} \frac{1}{k} - [\ln(4n) - \ln 1]\right)$

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

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

Since $\ln 1 = 0$, $\ln(2n) = \ln 2 + \ln n$, and $\ln(4n) = \ln 4 + \ln n = 2\ln 2 + \ln n$, we have

$s_{3n} = 2\ln 2 + \ln n + \displaystyle \left( \sum_{k=1}^{4n} \frac{1}{k} - \ln 4n \right)$

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

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

or

$s_{3n}= \displaystyle \frac{3}{2}\ln 2 + \displaystyle \left( \sum_{k=1}^{4n} \frac{1}{k} - \ln 4n \right) - \frac{1}{2}\left( \sum_{k=1}^{2n} \frac{1}{k} - \ln 2n\right) - \frac{1}{2}\left( \sum_{k=1}^{n} \frac{1}{k} - \ln n\right)$.

I now take the limit as $m \to \infty$:

$\displaystyle \lim_{n \to \infty} s_{3n} = \displaystyle \frac{3}{2}\ln 2 + \lim_{n \to \infty} \left( \sum_{k=1}^{4n} \frac{1}{k} - \ln 4n \right) - \frac{1}{2} \lim_{n \to \infty} \left( \sum_{k=1}^{2n} \frac{1}{k} - \ln 2n\right) - \frac{1}{2} \lim_{n \to \infty} \left( \sum_{k=1}^{n} \frac{1}{k} - \ln n\right)$.

This step reveals why I added and subtracted the integrals above: those gymnastics were necessary in order to reach a limit that converges.

As shown earlier in this series, if

$\displaystyle \lim_{m \to \infty} \left( \sum_{k=1}^m \frac{1}{k} - \ln m \right) = \gamma$,

the Euler-Mascheroni constant. Therefore, since the limit of any subsequence must converge to the same limit, we have

Applying these above, we conclude

$\displaystyle \lim_{m \to \infty} s_{3n} = \displaystyle \frac{3}{2}\ln 2 + \gamma - \frac{1}{2}\gamma - \frac{1}{2} \gamma = \displaystyle \frac{3}{2} \ln 2$,

which is different than $\ln 2$.

Technically, I’ve only shown so far that the limit of partial sums 3, 6, 9, … is $\displaystyle\frac{3}{2} \ln 2$. For the other partial sums, I note that

$\displaystyle \lim_{n \to \infty} t_{3n+1} = \displaystyle \lim_{n \to \infty} \left(s_{3n} + \displaystyle \frac{1}{4n+1} \right) = \displaystyle \frac{3}{2} \ln 2 + 0 = \displaystyle \frac{3}{2} \ln 2$

and

$\displaystyle \lim_{n \to \infty} t_{3n-1} = \displaystyle \lim_{n \to \infty} \left(s_{3n} - \displaystyle \frac{1}{2n} \right) = \displaystyle \frac{3}{2} \ln 2 - 0 = \displaystyle \frac{3}{2} \ln 2$.

Therefore, I can safely conclude that

$T = \displaystyle \lim_{n \to \infty} t_n = \displaystyle \frac{3}{2} \ln 2$,

which is different than the original sum $S$.