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i need help solving this task, if anyone had a similar problem it would help me a lot.

The task is : Find the sum of order

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

I'm not asking anyone to do my task, I need a hint on how to start, because I don't know how to start

LogicNotFound
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  • I don't need convergence, I just need the sum of this order. – LogicNotFound Jul 25 '21 at 12:32
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    “Order”? You mean just the regular sum of the whole thing, right? – FShrike Jul 25 '21 at 12:33
  • Yes, I mean the regular sum of the order. – LogicNotFound Jul 25 '21 at 12:35
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    I have absolutely no idea what “order” means; I’ve never seen it used in the context of infinite series! Anyway - do the partial sums appear to converge to anything? – FShrike Jul 25 '21 at 12:36
  • You can use $\frac{1}{1+x}=1-x+x^2-x^3+\ldots$ and then use some combination of differentiation, integration, and multiplication and division by $x$ applied to both sides to get something like your series and then set $x=\frac12$ – Henry Jul 25 '21 at 12:41
  • You could use a partial fraction decomposition as well, instead of one of the answers excellent log expansion, if you find that easier – FShrike Jul 25 '21 at 12:58
  • Perhaps you mean the order of the sum? – saulspatz Jul 25 '21 at 13:32

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Hint:

Adding and subtracting one ($n+1=n+2-1$) yields $$\begin{align} \sum_{n=0}^{\infty}\frac{(-1)^n (n+1)}{(n+2)2^{n}}&=\sum^\infty_{n=1}\frac{(-1)^n}{2^n}-\sum^\infty_{n=1}\frac{(-1)^{n}}{(n+2)2^n} \end{align} $$ Let $x=-\frac12$.

  • The first term of the right hand side is of the form $$g(x)=\sum^\infty_{n=1}x^n$$

  • The second term on the right-hand side is of the form $$\begin{align}f(x)&=\sum^n_{n=1}\frac{x^n}{n+2}=x^{-2}\sum^\infty_{n=1}\frac{x^{n+2}}{n+2}\\ &=x^{-2}\Big(\sum^\infty_{n=1}\frac{x^n}{n} - x-\frac{x^2}{2}\Big)\\ &=x^{-2}\Big(-\log(1-x)-x-\frac{x^2}{2}\Big) \end{align}$$

Mittens
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  • When dividing the sum, why does the counter go from 1? – LogicNotFound Jul 25 '21 at 13:38
  • This is just standard re-indexing: $\sum^\infty_{n=1}\frac{x^{n+2}}{n+2}=\sum^\infty_{n=3}\frac{x^n}{n}$. The rest is just adding (and subtracting) terms to get the complete well known series $\sum^\infty_{n=1}\frac{x^n}{n}=-\log(1-x)$ – Mittens Jul 25 '21 at 16:33