Harmonic series and its alternating counterpart

Question

I can't understand intuitively why the series

$$ \sum^{\infty}_{k=1} \frac{1}{k} $$

diverges while its counterpart with only the alternating signs does the opposite (converges)

$$\sum^{\infty}_{k=1} \frac{(-1)^k}{k} $$

Can someone explain?

Answer 1

This is a classic result. I am sure this is answered somewhere else.

My math professor's proof, I think is very intuitive:

$ \frac{1}{1} + \frac{1}{2} + \big(\frac{1}{3} + \frac{1}{4} \big)+ \big(\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}\big) + ... > \frac{1}{1} + \frac{1}{2} + \big(\frac{1}{4} + \frac{1}{4} \big)+ \big(\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}\big) + ...$

$ \Rightarrow \frac{1}{1} + \frac{1}{2} + \big(\frac{1}{3} + \frac{1}{4} \big)+ \big(\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}\big) + ... > \frac{1}{1} + \frac{1}{2} + \big(\frac{1}{2} \big)+ \big(\frac{1}{2} \big) + ...$

The right hand side clearly diverges, and thus the original sum diverges. You are adding smaller and smaller pieces, but the sum of an increasing number of those pieces always remains finite (sum of $2^n$ terms will always give you 1/2).

The alternating counterpart can be shown to be bounded using a similar argument: $ \frac{1}{1} - \big(\frac{1}{2} - \frac{1}{3} \big) - \big(\frac{1}{4} - \frac{1}{5}\big) - \big(\frac{1}{6} - \frac{1}{7}\big) ... < \frac{1}{1} - \big(\frac{1}{2} - \frac{1}{2} \big) - \big(\frac{1}{4} - \frac{1}{4}\big) - \big(\frac{1}{6} - \frac{1}{6}\big) + ... = 1$

Thus, the alternating series is bound by 1. Here you are subtracting something and then immedietely adding something slightly smaller (the way its grouped above). The net subtraction amount keeps reducing at a rate that is equal to $O(1/n^2)$. The actual value of the alternating series can be obtained from the taylor series of ln(1+x) at x=1.

Hope this helps.