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I just started the series chapter and I come across some series that I don't know how should I resolve them. All of them have the same structure : $$ \sum_{n=0}^{\infty} (-1)^n ... $$ For example: $$ \sum_{n=0}^{\infty} (-1)^n \frac{n+1}{2^n} or $$ $$ \sum_{n=0}^{\infty} (-1)^n \frac{n^2}{2^n} $$ Some tips on how should i aproache this would be greatly appreciated.

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I edited my post. Sorry it's n from 0 to infinity. –  user137209 Apr 26 at 13:44
I suggest you change the upper value to $m$ –  Claude Leibovici Apr 26 at 13:45

1 Answer 1

up vote 3 down vote accepted

First, absorb $(-1)^n$ with $\displaystyle\frac1{2^n}$ to find

$$\sum_{n=0}^{\infty}(-1)^n\frac{n+1}{2^n}=\sum_{n=0}^{\infty}(n+1)\left(-\frac12\right)^n$$ $$=\sum_{n=0}^{\infty}(n-1)\left(-\frac12\right)^n+2\underbrace{\sum_{n=0}^{\infty}\left(-\frac12\right)^n}_{\text{ Geomtric Series}}$$

For $|r|<1,$ $$\sum_{n=0}^{\infty}(n-1)r^n=\sum_{n=0}^{\infty}(n-1)r^n=\frac{d \sum_{n=0}^{\infty}r^n}{dr}$$

See also: Elementary problems and this

Can you manage the last problem from here?

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It is funny ! Today, you wrote at least twice $Geomtric$ $Series$. Cheers. –  Claude Leibovici Apr 26 at 13:48
Yes, i will manage the second problem. Thank you very much! –  user137209 Apr 26 at 13:50
@ClaudeLeibovici, Sorry for the repetition –  lab bhattacharjee Apr 26 at 13:51
@user137209, We can utilize en.wikipedia.org/wiki/… for the problem I've answered –  lab bhattacharjee Apr 26 at 13:52
Don't be sorry, please ! I just found funny that, being almost blind, I noticed that. At least, it proves that I read what you write (so nicely) ! Cheers. –  Claude Leibovici Apr 26 at 13:53

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