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Prove that $\sum_{n=0}^{\infty} \frac{n}{n+1}x^{n+1}=\frac{x}{1-x} + \ln(1-x)$

Right off the bat, I noticed that $\frac{x}{1-x}=x\frac{1}{1-x}= x \sum_{n=0}^{\infty}x^n$, and I know that $\ln(1+x)=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^n}{n}$. If somehow I can express $\ln(1-x)$ in terms of $\ln(1+x)$, then I'm set. Can I do something like $\ln(1-x)=\ln(1+(-x))=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{(-x)^n}{n}$?

What's next? Should I add these 2 together? Can someone give me an exact answer? Also, the radius of convergence is supposed to be $(-1, 1)$ right?

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    $\begingroup$ 1. That is true only for $|x| < 1$. 2. Did you know that $\frac{1}{1-x} = 1 + x + x^2 + ...$? $\endgroup$ – Kaind Apr 8 at 1:33
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    $\begingroup$ You are on the right track. For a formal proof, you'll have to justify that you are adding two power series term by term. $\endgroup$ – Hans Engler Apr 8 at 1:37
  • $\begingroup$ Call that power series $f(x)$. Take a derivative, factor an x and integrate. $\endgroup$ – Neutral Element Apr 8 at 1:40
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$$ \frac{n}{n+1}=1-\frac{1}{n+1} $$ so $$f(x)=\sum_{n\geq0}\frac{n}{n+1}x^n=\sum_{n\geq0}x^n-\sum_{n\geq0}\frac{x^n}{n+1}$$ Of course both the two power series that appear above converge for all $x\in(-1,1)$. Can you finish from here?

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Know: \begin{equation} \frac{1}{1-x} = \sum_{n=0}^{\infty}x^{n} \implies \frac{1}{1-x} - 1 = \sum_{n=1}^{\infty}x^{n}, \label{first series} \end{equation} which holds for $|x| < 1$, and \begin{equation} \log(1+x) = -\sum_{n=1}^{\infty}\frac{1}{n}(-x)^{n} \implies \log(1-x) = -\sum_{n=1}^{\infty}\frac{1}{n}x^{n}, \label{second series} \end{equation} also valid for $|x| < 1$. If we now look at our series of interest, \begin{equation} f(x) = \sum_{n=0}^{\infty}\frac{n}{n+1}x^{n+1}, \end{equation} we can use an index substitution $k = n+1 \iff n = k-1$ so that \begin{align} f(x) &= \sum_{k=1}^{\infty}\frac{k-1}{k}x^{k}\\ &= \sum_{k=1}^{\infty}\left(1-\frac{1}{k}\right)x^{k},\\ \end{align} and under the condition that our series converges, we can use the linearity of the series operator to show \begin{equation} f(x) = \sum_{k=1}^{\infty}x^{k} + \left(-\sum_{k=1}^{\infty}\frac{1}{k}x^{k}\right), \end{equation} provided both the individual series converge (which they do). Notice that on the right-hand side of the above equation, the leftmost series is equal to $1/(1-x) - 1$ (from \eqref{first series}), and the rightmost series is equal to $\log(1-x)$ (from \eqref{second series}). Therefore \begin{align} f(x) &= \frac{1}{1-x} - 1 + \log(1-x)\\ &= \frac{x}{1-x} + \log(1-x). \end{align}

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\begin{equation} f(x) = \sum_{n=0}^{\infty}\frac{n}{n+1}x^{n+1} \end{equation} $$f'(x)=\sum_{n=0}^{\infty} n x^n=x\sum_{n=0}^{\infty} n x^{n-1}=x\left(\sum_{n=0}^{\infty} x^{n}\right)'=\frac{x}{(x-1)^2}$$ $$f'(x)=\frac{x}{(x-1)^2}=\frac{x-1+1}{(x-1)^2}=\frac 1{x-1}+\frac 1 {(x-1)^2}$$ Just integrate.

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