# Rearrangements of the harmonic series

Trying to understand how rearrangements work. A very common example of rearrangements seems to be the alternating harmonic series,

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

Plugging in values of $n$ gives,

$$1-\frac{1}{2}+\frac {1}{3}-....$$ and so on. How can I rearrange this sum so that the first $10$ sum to $0$.

It seems that I must alter it in some way grouping together positive and negative terms so that the first 10 terms sum to 0.

• This video may be instructive . Anyway, there is no trivial reason to conclude that such an arrangment exists. You can arrange the summand in order to get that the infinite series is $0$, but this does not mean that you can find some arrangement where the first 10 terms add up to $0$. – Crostul Feb 2 '17 at 21:42

Playing around with Egyptian fractions yields $$\frac 1{15}+\frac 1{17}+\frac 1{19}+\frac 1{21}+\frac 1{25}-\left(\frac 14 + \frac 1{64} + \frac 1{8634} + \frac 1{679007792} + \frac 1{662761647866270400}\right)=0$$
This is a classic example of a conditional convergent series. So when you unravel the terms by positives and negatives, we essentially get two separate divergent (!) series. One diverges to infinity and the other to minus infinity. So by choosing the terms not alternating, but by "choice", you can make the series converge to any sum you want. In your question you are at bit "ambiguous" by saying that you want the first $10$ terms sum to zero. Obviously it will never exactly be equal to zero (Crostul's comment!). But you can start of with $1$ and then pick $-1/2$ and then pick $-1/4$, then pick $-1/6$, then pick $-1/8$ so that you have dipped below $0$. Now pick $1/3$ which makes your partial sum go above $0$. Then pick $-1/10$ etc. When continuing in this fashion, you can get as close to $0$ as you wish...