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

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?

• 1. That is true only for $|x| < 1$. 2. Did you know that $\frac{1}{1-x} = 1 + x + x^2 + ...$? – Kaind Apr 8 at 1:33
• 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. – Hans Engler Apr 8 at 1:37
• Call that power series $f(x)$. Take a derivative, factor an x and integrate. – Neutral Element Apr 8 at 1:40

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