Slick proofs that if $\sum\limits_{k=1}^\infty \frac{a_k}{k}$ converges then $\lim\limits_{n\to\infty} \frac{1}{n}\sum\limits_{k=1}^n a_k=0$

Question

I'm looking for slick proofs that if $a_n$ is a sequence of complex numbers such that $\sum\limits_{k=1}^\infty \frac{a_k}{k}$ converges then $\lim\limits_{n\to\infty} \frac{1}{n}\sum\limits_{k=1}^n a_k=0$.

My not so slick proof:

Let $A_x=\sum\limits_{k=1}^{x} a_k$ and apply an Abel sum with the function $f(x)=\frac{1}{x}$. We get $$\sum\limits_{k=1}^n \frac{a_k}{k}-\int\limits_1^n\frac{A_x}{x^2}dx=\frac{A_n}{n}.$$ Of course we can split the integral into chunks to get $$\sum\limits_{k=1}^n \frac{a_k}{k}-\sum\limits_{i=1}^n \left( \frac{1}{k+1}-\frac{1}{k} \right) A_k=\frac{A_n}{n}.$$ If we expand $A_k$ on the left side, it telescopes and yields $$\sum\limits_{k=1}^n \frac{a_k}{k} + \sum\limits_{k=1}^n \left( 1-\frac{1}{k+1}\right) a_k=\frac{A_n}{n}\iff \frac{(n-1)A_n}{n}=\sum\limits_{k=1}^n\frac{a_k}{k(k+1)}.$$ The norm of the right side of the equation is clearly $\mathcal O(\log(n))$ (using the triangle inequality and the fact that $\frac{a_k}{k}$ is bounded). The desired result follows after dividing both sides by $n-1$.

I would also appreciate some proof verification. Initially, I thought that the problem was going to be really easy, but it took me a bit of effort. Am I missing something major? The original problem is for real sequences, but I don't think this helps. Obviously if the sequence converged absolutely then it would also be absolutely trivial.

Answer 1

Let $b_k = \dfrac{a_k}{k}$, and define

$$R_m = \sum_{k = m}^{\infty} b_k.$$

Then

\begin{align} \frac{1}{n} \sum_{k = 1}^n a_k &= \frac{1}{n}\sum_{k = 1}^n kb_k \\ &= \frac{1}{n} \sum_{k = 1}^n\sum_{m = k}^n b_m \\ &= \frac{1}{n} \sum_{k = 1}^n \bigl(R_k - R_{n+1}\bigr) \\ &= \Biggl(\frac{1}{n} \sum_{k = 1}^n R_k\Biggr) - R_{n+1}. \end{align}

Now use that $R_k \to 0$, and that the Cesàro means of a convergent sequence converge to the same limit.