Thoughts on Infinity (Part 3d)

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.

In yesterday’s post, I showed that

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

This can be (sort of) confirmed using commonly used technology — in particular, Microsoft Excel. In the spreadsheet below, I typed:

• 1 in cell A1
• =POWER(-1,A1-1)/A1 in cell B1
• =B1 in cell C1
• =A1+1 in cell A2
• =POWER(-1,A2-1)/A2 in cell B2
• =C1+B2 in cell C2
• Then I used the FILL DOWN command to fill in the remaining rows. Using these commands cell C10 shows the sum of all the entries in cells B1 through B10, so that

$1 - \displaystyle \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} - \frac{1}{8} + \frac{1}{9} - \frac{1}{10} \approx 0.645634921$

Filling down to additional rows demonstrates that the sum converges to $\ln 2$, albeit very slowly (as is typical for conditionally convergent series). Here’s the sum up to 200 terms… the entry in column E is the first few digits in the decimal expansion of $\ln 2$.

Here’s the result after 2000 terms:

20,000 terms:

And finally, 200,000 terms. (It takes a few minutes for Microsoft Excel to scroll this far.)

We see that, as expected, the partial sums are converging to $\ln 2$, as expected. Unfortunately, the convergence is extremely slow — we have to compute 10 times as many terms in order to get one extra digit in the final answer.

1. howardat58

/  October 27, 2015

I suggest adding two adjacent terms at a time. This will show the convergence faster, and get rid of most of the oscillation.

• John Quintanilla

/  October 27, 2015

I don’t disagree; however, my primary point was to illustrate how slowly the unaltered series converges, as opposed to techniques for accelerating the convergence.