Determining precisely where $\sum_{n=0}^\infty\frac{z^{n+2}}{(n+1)(n+2)}$ converges at the boundary of disk of convergence?

Question

I'm trying to determine the behaviour on the boundary of the disk of convergence of the series $\sum_{n=0}^\infty\frac{z^{n+2}}{(n+1)(n+2)}$ for $z\in\mathbb{C}$.

The radius of convergence $R$ is $$R= \lim_{n\to\infty} \frac{|a_n|}{|a_{n+1}|} = \lim_{n\to\infty} \frac{\frac{1}{(n+1)(n+2)}}{\frac{1}{(n+2)(n+3)}} = \lim_{n\to\infty} \frac{(n+2)(n+3)}{(n+1)(n+2)} = \lim_{n\to\infty} \frac{n+3}{n+1} = 1$$

Now, I want to study what happens when $|z| = 1$. We want to use Dirichlet's test:

1. $a_n \geq a_{n+1} \iff \frac{1}{(n+1)(n+2)} \geq \frac{1}{(n+2)(n+3)}$, which is true.
2. $\lim_{n\to\infty} a_n = 0 \iff \lim_{n\to\infty} \frac{1}{(n+1)(n+2)} = 0$, which is true.
3. There exists $M>0$ such that $\left|\sum_{n=1}^Nb_n\right|\leq M$ for every $N\in\mathbb{N}$:

$$\sum_{n=0}^N b_n = \sum_{n=0}^N z^{n+2} = \frac{z^2-z^{k+3}}{1-z}$$ Then, $$\left|\sum_{n=0}^N z^{n+2}\right| = \left|\frac{z^2-z^{k+3}}{1-z}\right| \leq \frac{2}{|1-z|}$$

The third condition is satisfied for every $z \neq 1$ in the circle.

The problem is that this result does not match the solution that our teacher provided us. Any help would be greatly appreciated because I'm stuck at this problem!

Answer 1

Let $f(z)=\sum_{n=0}^\infty\frac{z^{n+2}}{(n+1)(n+2)}$. $$ \frac{d}{dz}f(z)=\sum_{n=0}^\infty \frac{z^{n+1}}{n+1}. $$ and $$ \frac{d^2}{dz^2}f(z)=\sum_{n=0}^\infty z^n=\frac1{1-z}. $$ Since $f'(0)=0$, we get $$ f'(z)=\int_0^z \frac{du}{1-u}=-\ln(1-z). $$ Finally, since $f(0)=0$, we get $$ f(z)=-\int_0^z \ln(1-u)du=z+(1-z)\ln(1-z). $$ The series obviously converges for $|z|<1$. See this.

On the boundary $|z|=1$ the function remains analytic except at $z=1$, where there is a branch point. The question then is whether convergence continues to hold at $z=1$.

A branch point allows convergence if all branches of the function converge to a common limit at the branch point. But Plugging in $w=1-z$ we have

$w\ln w = w(\ln|w|+i\arg w)$

$=|w|\exp(i\arg w)\ln|w| + iw\arg w$

From real variable analysis we know that as $w\to0$, $|w|\ln|w|\to0$; and $\arg w$ remains finite on every branch of the complex logarithm. Therefore

$w\ln w =(1-z)\ln (1-z)\to0$

on all branches of the logarithm as $z\to1, w=1-z\to0$, which means the original series converges at $z=1$ Thus completing convergence on all of the boundary $|z|=1$.