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.)

log2series5We 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.



4 thoughts on “Thoughts on Infinity (Part 3d)

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