# Evaluate $\lim_{x\to 1}\sum_{1}^ \infty \frac{x^n}{n(n+1)}$

Let

$$f(x) = \sum_{n=1}^ \infty \frac{x^n}{n(n+1)}$$

I'm trying to evaluate

$$\lim_{x\to 1} \, f(x)$$

What I've done so far is getting the radius of convergence, which is$\ -1\le x \le 1$ using the ratio test.

I'm wondering what has to be done from here, does the ratio test imply anything about the limit I'm being asked for?

Hints would be greatly appreciated.

• Abel theorem, or you can sum the series too (for this see what would be $(x^3f'(x))'$). – user568221 Jun 8 '18 at 20:13

Several comments on this are worthwhile.

In the radius of convergence, $$f(x)=\sum_{n\ge 1}(\frac{x^n}{n}-\frac{x^{n}}{n+1})=\sum_{n\ge 1}\frac{x^n}{n}-\frac{1}{x}\sum_{n\ge 2}\frac{x^n}{n}=(\frac{1}{x}-1)\ln(1-x)+1,$$where we have used$$\sum_{n\ge 1}\frac{x^n}{n}=-\ln(1-x).$$Our telescoping sum implies $f(1)=1$, which is consistent with the above since $$\lim_{y:=1-x\to 0^+}y\ln y=0.$$

Your sum diverges by the ratio test if $|x|>1$, so a two-sided $x\to 1$ limit does not exist, although an $x\to 1^-$ left-handed limit does.

If you try to redefine $f$ by analytic continuation, the above log-based expression requires complex numbers because $\Im\ln(1-x)=\pi i$ for $x>1$ (although not if $x<-1$), provided the principal natural logarithm is taken. But since $\lim_{x\to 1}(\frac{1}{x}-1)$, the two-sided limit exists in this case.

Assuming $|x| < 1$

$$\sum_{k=0}^{\infty}x^k = \frac{1}{1-x}$$

$$\int \sum_{k=0}^{\infty}x^k dx = \sum_{k=1}^{\infty}\frac{x^{k+1}}{k+1} = -\log(1-x)$$

$$\sum_{k=1}^{\infty}\frac{x^k}{k(k+1)} = \frac{1}{x}\int \sum_{k=1}^{\infty}\frac{x^{k+1}}{k+1} = \sum_{k=1}^{\infty}\frac{x^k}{k(k+1)} = \frac{x-x \log (1-x)+\log (1-x)}{x}$$

as can be verified easily

$$\lim_{x\to 0}\frac{x-x \log (1-x)+\log (1-x)}{x} = 0$$