Prove that the series converges absolutely for all $z \in \mathbb C$ with $|z|=1$

Question

Prove that the series converges absolutely for all $z \in \mathbb C$ with $|z|=1$.

$\sum_{k=2}^{\infty}\frac{z^k}{k(k-1)}$

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$\sum_{k=2}^{\infty}\frac{z^k}{k(k-1)} = \sum_{k=2}^{\infty}\frac{1}{k(k-1)}*z^k$

For the root test and ratio test I get $1$ as a result, which doesn't help at all.

Using the Leibniz criterion I only get the result that it converges but it doesn't tell me anything about absolute convergence.

By definition a series $\sum a_k$ converges absolutely if $\sum |a_k|$ converges.

$|z| = 1$ seems to be exactly on the radius of convergence.

So how do I prove it, if the usual criteria don't work?

Answer 1

We just check it directly. If $|z| = 1$ then $|\frac{z^k}{k (k - 1)}| = \frac{1}{k (k - 1)} = \frac{1}{k - 1} - \frac{1}{k}$.

Then we see that $\sum\limits_{k = 2}^\infty |\frac{z^k}{k (k - 1)}| = \sum\limits_{k = 2}^\infty \frac{1}{k - 1} - \frac{1}{k} = 1$ by the telescoping series test.

Answer 2

Can't you apply the fact: $$\frac{1}{k(k-1)} \leq \frac{1}{(k-1)^2}.$$

Since $\sum\frac{1}{k^2}$ is convergent, the given series should also converge absolutely for $|z|=1$.