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I have the series $\sum_{n=0}^{\infty}\frac{n}{n+1}x^n$. It is convergent for $-1<x<1$.

I want to find the sum function. I've tried the following: \begin{align*} \sum_{n=0}^{\infty} x^n &\sim \frac{1}{1-x}\\ \sum_{n=0}^{\infty} \frac{x^{n+1}}{1+n}&\sim -\log(1-x)&\text{Integrating both sides}\\ \sum_{n=0}^{\infty} \frac{x^{n}}{1+n}&\sim-\frac{\log(1-x)}{x}&\text{Divide with $x$}\\ \sum_{n=0}^{\infty}\frac{nx^{n-1}}{1+n}&\sim\frac{1}{(1-x)x}+\frac{\log(1-x)}{x^2}&\text{Derivative with respect to $x$}\\ \sum_{n=0}^{\infty}\frac{nx^{n}}{1+n}&\sim\frac{1}{(1-x)}+\frac{\log(1-x)}{x}&\text{Multiply by $x$} \end{align*}

I can't decide on what to do from this point. I seem to miss the $n$ in the nominator, but changing index does not really help med out. Can you point me in the right direction?

Edit: I used the tips and finished it. Thank you.

bubububub
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1 Answers1

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If your function is $f(x)$, then $xf(x)=\sum_{n=1}^\infty\frac n{n+1}x^{n+1}$ and therefore its derivative is$$\sum_{n=1}^\infty nx^n=x\sum_{n=1}^\infty nx^{n-1}=x\left(\sum_{n=1}^\infty x^n\right)'.$$Can you take it from here?

José Carlos Santos
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