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 nth 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.

Leave a comment

1 Comment

  1. Thoughts on Infinity: Index | Mean Green Math

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: