If $\sum_{n=1}^\infty a_n$ converges, prove that $\lim_{n\to \infty} (1/n) \sum_{k=1}^n ka_k = 0$. the series $a_n$ is any arbitrary converging series.
My thought process was that the $1/n$ will definitely go to zero as n approaches infinity; however, the series $k*a_k$ seems to approach infinity at the same rate. This confuses me because if both the numerator and denominator approach infinity, I think that the overall equation will equal 1, not zero.
I also considered that the sum is geometric, but I do not believe that is the case, because both $a_k$ and $k$ are variables that are not constant. I was thinking that I could argue if the series $a_k$ converges to a point, eventually the terms of the sequence will be so close to each other that it resembles a constant multiplier, but that argument is not very strong.
Any tips, hints, or leads at the solution are appreciated!
 A: Let $x_n=\sum_{k=1}^n a_k$. The hypothesis is that $x_n$ converges to some $x$. Note that $\sum_{i=1}^n x_i = \sum_{k=1}^n (n-k+1)a_k$, and recall (or if necessary, prove) that if $y_n\to y$ then $$\hat y_n=\frac{1}{n}\sum_{i=1}^n y_i\to y$$
A: Hint: from the convergence of $\sum_{n\geq 1}a_n$, we know that the sequence $\{a_n\}_{n\geq 1}$ has bounded partial sums. We have:
$$ \frac{1}{n}\sum_{k=1}^{n} k\,a_k = \sum_{k=1}^{n}\frac{k}{n}\,a_k = \sum_{k=1}^{n} a_k - \sum_{k=1}^{n}\left(1-\frac{k}{n}\right)a_k. $$
Now apply Dirichlet's criterion to the last sum regarded as a series.
Once you know that $\frac{1}{n}\sum_{k=1}^{n} k\,a_k$ is convergent, convergence to zero follows from the dominated convergence theorem.
A: Let $S_k = \sum_{j=1}^{k}a_j.$ We are given $S_k \to$ some $L.$ Summing by parts, we get
$$\sum_{k=1}^{n}ka_k = nS_n - \sum_{k=1}^{n-1}S_k.$$
Dividing by $n$ gives $S_n$ minus $(n-1)/n$ times the $(n-1)$st Cesaro mean of the $S_k,$ which also converges to $L.$ Our limit is therefore $L-L=0.$
