# Thoughts on Infinity (Part 3c)

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.

Define

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

By the alternating series test, this series converges. However,

$\displaystyle \sum_{n=1}^\infty |a_n| = \displaystyle \sum_{n=1} \frac{1}{n}$,

which is the divergent harmonic series which was discussed earlier in this series. Therefore, the series $S$ converges conditionally and not absolutely.

To calculate the value of $S$, let $s_n = \displaystyle \sum_{k=1}^n \frac{(-1)^{k-1}}{k}$, the $n$th partial sum of $S$. Since the series converges, we know that $\displaystyle \lim_{n \to \infty} s_n$ converges. Furthermore, the limit of any subsequence, like $\displaystyle \lim_{n \to \infty} s_{2n}$, must also converge to $S$.

If $n$ is even, so that $n = 2m$ and $m$ is an integer, then

$s_{2m} = \displaystyle \sum_{k=1}^n \frac{(-1)^{k-1}}{k}$

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

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

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

$= \displaystyle \sum_{k=1}^{2m} \frac{1}{k} - \sum_{k=1}^m \frac{1}{k}$.

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

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

or

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

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

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

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

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

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

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

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

In yesterday’s post, I showed that if

$t_m = \displaystyle \left( \sum_{k=1}^m \frac{1}{k} - \ln m \right)$

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

the Euler-Mascheroni constant. Therefore, the limit of any subsequence must converge to the same limit; in particular,

$\displaystyle \lim_{m \to \infty} t_{2m} =\displaystyle \lim_{m \to \infty} \left( \sum_{k=1}^{2m} \frac{1}{k} - \ln 2m \right)= \gamma$.

Applying these above, we conclude

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

or

$S = \ln 2$.