Question

Consider the series \[\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\] and let \[A\in \mathbb{R}\] be any real number. Show that by rearranging the terms of the series, the sum will be a number in the interval \[(A-1, A+1).\]

Answer 1

I think this series converges since each term is alternating in sign, and we have \[|u_{n+1}| < |u_{n}| \forall n\] and \[\lim_{n\rightarrow \infty}u_n = 0\] and we may use the alternating series theorem to deduce convergence. (here I've used \[u_n\] to denote the nth term of the series.
The trouble is, this converges right, so that means the limit exists and exists in some specific interval (A-1, A+1) where A isn't just any number...maybe I'm confused :/

Answer 2

any condidionally convergent series can be rearranged so that the sum is any number you want. That is the case for thise series.

Answer 3

*conditionally

Answer 4

oh yeah! The Riemann Series Theorem, thanks Zarkon!

Answer 5

Here are two proofs:

http://pirate.shu.edu/~wachsmut/ira/numser/proofs/sumorder.html

and

(PDF WARNING)
http://www.csun.edu/~hcmth017/riemann1/riemann1.pdf

It's a wild result, and it's why we always must check if we have abs vs conditional convergence before jumping to conclusions. Even the great Ramanujan fell victim to this, he ended up with 1 + 2 + 3 + ... = -1/12 because of this. :-p

Answer 6

Amazing, thanks for all your help!