Use power series $\sum^{\infty}_{n=1} \frac {z^{n+1}} {n(n+1)}$ to show $\sum^{\infty}_{n=1} \frac {1} {n(n+1)} =1$.

Question

Consider $\sum^{\infty}_{n=1} \frac {z^{n+1}} {n(n+1)}$ (power series). I've found that the sum-function $g(z) := \sum^{\infty}_{n=1} \frac {z^{n+1}} {n(n+1)}$ is defined and continuous on $|z| \le 1$.

Let $f$ be the restriction of $g$ to $[-1,1]$. I've shown that $f(x) = (1-x)\log(1-x)+x$.

How can I use these results to show $\sum^{\infty}_{n=1} \frac {1} {n(n+1)} =1$ ?

I'm well aware that $\frac 1 n - \frac 1 {n+1} = \frac 1 {n(n+1)}$, but I think I should use what I've proved instead of looking at the partial sums.

Does it hold to say $\lim_{x \rightarrow 1} f(x) = \lim_{x \rightarrow 1} (1-x)\log(1-x)+x$ ? I know, since $f$ is continuous $\lim_{x \rightarrow 1} f(x) = f(1)$, but $(1-x)\log(1-x)+x$ is not defined for $x=1$ ?

Answer 1

Hint

When $x$ goes to $0$, $x\log(x)$ has a limit of $0$ and, so, when $x$ goes to $1$, $(1-x)\log(1-x)$ has also a limit of $0$.

Answer 2

Yes, the important missing ingredient is Abel's Theorem. This is also used, for example, to prove that $$\sum_{n=1}^\infty (-1)^{n+1}\frac1n = \log 2.$$