TITLE: Question about convergence of series if $\{n a_n\} \to 0$
QUESTION [5 upvotes]: This is part of Rudin's PMA Exercise 3.14 (d). If I understand correctly, it would be helpful to prove the following:

Let $a_n$ be some sequence. Assume that $\lim_{n\to\infty} na_n = 0$. 

Prove that
$$\lim_{n\to\infty} \sum_{k=1}^{n} \frac{k\,a_k}{n+1} = 0.$$

REPLY [4 votes]: Let $\epsilon>0.$ There is some $n_0 \in \mathbb{N}$ such that $|na_n|< \epsilon $ for all $n>n_0.$ Then $$ \bigg|\sum_{k=1}^n ka_k \bigg|= \bigg|\sum_{k=1}^{n_0} ka_k + \sum_{k=n_0+1}^n ka_k\bigg| < \bigg|\sum_{k=1}^{n_0}ka_k\bigg| +(n-n_0)\epsilon$$
which leads to $$\limsup_{n\to \infty} \bigg|\sum_{k=1}^n \dfrac{ka_k}{n+1}\bigg| \leq\epsilon.$$
Since this is true for any $\epsilon>0$ we must have  $$\lim_{n\to \infty} \sum_{k=1}^n \dfrac{ka_k}{n+1}=0.$$