# How to determine a variable inside the lower and upper bound of a sum inside a infinite sum?

I hope you can help me on a problem I'm working on a long time.

We had to rearrange the alternating harmonic series

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

in such a way, that we get 5.

I split the series up in two new ones. One with the positive numbers $\Bigl(\sum\frac{1}{2i-1}\Bigl)$ and one with the negative numbers $\Bigl(\sum-\frac{1}{2t}\Bigl)$. I also wrote it down in a short formula but I don't know how the determine $\alpha$:

$$\sum_{n=1}^{\infty}\Biggl(\sum_{i=(n-1)\alpha+1}^{(n-1)\alpha+\alpha} \frac{1}{2i-1} +\sum_{t=n}^{n} -\frac{1}{2t}\Biggl)=5$$

What is the easiest way to determine $\alpha$?

(My idea was to add positive numbers until I'm over 5, then subtract a negative number so we are below 5, add positive until I'm above 5 again, subtract a negative and so on.

• take a look here – Masacroso Nov 25 '17 at 19:36
• Your idea is precisely a proof sketch of Riemann's theorem. Whether there's a more explicit way to write down the rearrangement, I don't know. – spaceisdarkgreen Nov 25 '17 at 19:38
• @ThePirateBay you can rearrange conditionally convergent series to get any value you want. en.wikipedia.org/wiki/Riemann_series_theorem – spaceisdarkgreen Nov 25 '17 at 19:39
• @ThePirateBay We have to rearrange the first series in such a way, that we get 5. I know the riemann theorem and i know how to rearrange, that it will become 1/2 log2 but that doesnt help... – Matthias N. Nov 25 '17 at 19:40
• @spaceisdarkgreen we dont have to proof Riemanns Theorem, we have to rearrange in a way, that it converges to exactly 5. – Matthias N. Nov 25 '17 at 19:43