# Infinite sum with alternated terms

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.

• Do you mean $\sum_{n = 0}^\infty (n+1)\frac{(-2)^n}{\pi^{n-1}}$? – guidoar Oct 30 '18 at 23:32
• Yes, is a typing error, thanks – Ezequiel Saidman Oct 30 '18 at 23:34

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!

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

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 $$(n-1)+2$$.