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Im stuck calculating this infinite sum

$\sum_{i=0}^{\infty} (i+1)\frac{(-2)^i}{\pi^{i-1}}$

I know what the series converges because the limit test is conclusive, but I need to calculate the sum and I don't know how.

I would appreciate if you give me a help.

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    $\begingroup$ Do you mean $\sum_{n = 0}^\infty (n+1)\frac{(-2)^n}{\pi^{n-1}}$? $\endgroup$ – Guido A. Oct 30 '18 at 23:32
  • $\begingroup$ Yes, is a typing error, thanks $\endgroup$ – Ezequiel Saidman Oct 30 '18 at 23:34
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I will give you a hint and try to work it out yourself! :)

So, try to rewrite the sum to

$$ \sum_{n=0}^{\infty} (n+1)\cdot a^{n} $$ for a certain $a$ and now you will need to apply the following trick: $$ \sum_{n=0}^{\infty} (n+1)\cdot a^{n} = \sum_{n=0}^{\infty} \frac{d}{d a} a^{n+1} = \frac{d}{d a} \sum_{n=0}^{\infty} a^{n+1} = \frac{d}{d a}\frac{a}{1-a} = \frac{1}{(1-a)^2}. $$ In general applying this trick multiple times can help you solve any sum of the form $$ \sum_{n=0}^{\infty} p(n) a^n $$ for a polynomial $p(n) = p_0 + p_1 n + p_2 n^2 + \dots + p_m n^m$.

Edit: NB: Note that for this to work we need $|a|<1$. This thus does not mean that $a$ cannot be negative.

I hope this helps, if you have any questions feel free to comment!

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Rewriting the sum we get:

$$ \sum_{n \geq 0}(n+1)\frac{(-2)^n}{\pi^{n-1}} = \sum_{n \geq 1}n\frac{(-2)^{n-1}}{\pi^{n-2}} = \pi\sum_{n \geq 0}n\left(\frac{-2}{\pi}\right)^{n-1}. $$

Now, it is well known that when $z \in B_1(0) \subseteq \mathbb{C}$,

$$ \sum_{n \geq 0}x^n = \frac{1}{1-x}, $$

and differentiating we have that

$$ \sum_{n \geq 1}nx^{n-1} = \frac{1}{(1-x)^2}. $$

Thus, $$ \sum_{n \geq 0}(n+1)\frac{(-2)^n}{\pi^{n-1}} = \frac{1}{\pi}\sum_{n \geq 0}n\left(\frac{-2}{\pi}\right)^{n-1} = \frac{\pi}{(1+\frac{2}{\pi})^2} = \frac{\pi^3}{(2+\pi)^2}. $$

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Hints: for $|t| <1$ we have $\sum nt^{n} = t \frac d {dt} \sum t^{n} =-\frac t {(1-t)^{2}}$. Take $t=\frac {-2} {\pi}$. Split $n+1$ into $(n-1)+2$.

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