If $g(x)=\sum_{n=1}^{\infty}\frac{x^n}{n(n+1)}$, then prove that $\lim\limits_{x\to 1^{-}}g(x)$ exists and find this limit

Question

Let $$g(x)=\sum_{n=1}^{\infty}\frac{x^n}{n(n+1)}.$$ Prove that $\lim\limits_{x\to 1^{-}}g(x)$ exists and find this limit.

As regards the interval of convergence of the series, here is what I have done:

$$\lim\limits_{n\to \infty}\Big|\frac{x^{n+1}}{(n+1)(n+2)}\frac{n(n+1)}{x^{n}}\Big|$$ $$=|x|\lim\limits_{n\to \infty}\Big|\frac{n}{n+2}\Big|=|x|$$ This implies that the series converges absolutely for values of $|x|<1$, diverges for values of $|x|>1$. When $x=1$, the series converges by Comparison test. When $x=-1$, the series converges by Alternating series test.

My question is, does this case: When $x=1$, the series converges by Comparison test, solve my problem? If not, can anyone help me with a the proof because I got stuck at this point. Thanks for your help!

Answer 1

Notice $g$ is a power series and is thus uniformly convergent within the radius of convergence. With a little work we can thus show that as we approach $1$ from below we can interchange the limit and summation to get

$$\begin{align} \lim_{x \to 1^-}g(x)&=\lim_{x \to 1^-}\sum_{n=1}^{\infty}\frac{x^n}{n(n+1)}\\ &=\sum_{n=1}^{\infty}\frac{1}{n(n+1)} =\lim_{k \to \infty}\sum_{n=1}^{k}\frac{1}{n(n+1)}\\ &=\lim_{k \to \infty}\sum_{n=1}^{k}\frac{1}{n(n+1)} = \lim_{k \to \infty}\sum_{n=1}^{k}\left(\frac{1}{n} - \frac{1}{n+1}\right)\\ \end{align}$$ Which is an easily evaluated, telescoping series.

Answer 2

An elementary calculation shows that for every $|x|<1$,

$$g(x)=\frac{x+\log(1-x)-x\log(1-x)}{x}$$

hence, by L'hopital's rule, $\lim_{x\to 1^{-}}g(x)=1$.

Answer 3

$$\sum_{n=1}^\infty \frac{x^n}{n(n+1)}=\sum_{n=1}^\infty x^n(\frac{1}{n}-\frac{1}{n+1})\\ =\sum_{n=1}^\infty\frac{x^n}{n}-\sum_{n=1}^\infty\frac{x^n}{n+1}=\sum_{n=1}^\infty\frac{x^n}{n}-\frac{1}{x}\sum_{n=1}^\infty\frac{x^{n+1}}{n+1}\\ =\sum_{n=1}^\infty\frac{x^n}{n}-\frac{1}{x}\sum_{n=2}^\infty\frac{x^{n}}{n}=\sum_{n=1}^\infty\frac{x^n}{n}-\frac{1}{x}\left(-x+\sum_{n=1}^\infty\frac{x^{n}}{n}\right) \\ =\left(1-\frac{1}{x}\right)\sum_{n=1}^\infty\frac{x^n}{n}+1=-\log(1-x)\left(\frac{x-1}{x}\right)+1\rightarrow1, x\rightarrow1^-$$