# How to prove that $\sum\limits_{k= 0}^n \frac{ku_k}{n^2}$ converges towards zero if $u_n \to 0$

I would like to prove that the following sequence converges to zero, given that $U_n$ converges to zero

$$v_n = \sum_{k= 1}^n \frac{kU_k}{n^2}$$

I have tried to use the epsilon definition and I got here :

$$- \epsilon \sum k <\sum_{k= N}^n \frac{kU_k}{n^2}< \epsilon \sum k$$

However I don't know what to do next, since this only work for $k \ge N$ where $N$ is the range that I got from the convergence epsilon definition for $u_n$

I there a better way using some theorem to prove this ? I was thinking about cesaro theorem .

• I can't understand how this would work ? could you please elaborate a bit more ? – Anis Souames Dec 2 '17 at 18:16
• This is a trivial consequence of the Stolz-Cesàro theorem. Let $a_n = \sum_{k=1}^{n}kU_k$ and $b_n = n^{2}$. Then, $\frac{a_{n+1}-a_n}{b_{n+1}-b_n} = \frac{(n+1)U_{n+1}}{2n+1} \to 0$ as $n\to \infty$. – Vinícius Novelli Dec 2 '17 at 18:19

Using the definition of limit, given an $\epsilon\gt0$, there is an $N$ so that $$n\gt N\implies|u_n|\lt\epsilon\tag1$$ Furthermore, find $$M=\max_{n\le N}|u_n|\tag2$$ Then, for $n\gt N$, \begin{align} \left|\,\sum_{k=1}^n\frac{ku_k}{n^2}\,\right| &\le\left|\,\sum_{k=1}^N\frac{ku_k}{n^2}\,\right|+\left|\,\sum_{k=N+1}^n\frac{ku_k}{n^2}\,\right|\\ &\le\underbrace{M\frac{N^2+N}{2n^2}}_{\to0}+\underbrace{\epsilon\frac{n^2+n}{2n^2}}_{\to\frac\epsilon2}\tag3 \end{align} Thus, $$\limsup_{n\to\infty}\left|\,\sum_{k=1}^n\frac{ku_k}{n^2}\,\right|\le\frac\epsilon2\tag4$$ Since this is true for all $\epsilon\gt0$, we have $$\lim_{n\to\infty}\left|\,\sum_{k=1}^n\frac{ku_k}{n^2}\,\right|=0\tag5$$
• $\frac {n^2+n}{2n^2}$ is greater than $\frac {1}{2}$, but otherwise, this seems fine. Just replace $\frac {\epsilon}{2}$ with $\epsilon$. – Michael L. Dec 2 '17 at 18:50
• The limit is $\frac12$, so we don't need to replace $\frac\epsilon2$. – robjohn Dec 2 '17 at 19:06
• So $M$ in (2) is the greatest term $u_n$ for which $n<=N$ ? – Anis Souames Dec 2 '17 at 19:08
• This is the maximum of $|u_n|$ for all $n\le N$. – robjohn Dec 2 '17 at 19:11
• Since $|u_k|\le\epsilon$ and $\sum\limits_{k=N+1}^n\frac{k}{n^2}\le\sum\limits_{k=1}^n\frac{k}{n^2}=\frac{n^2+n}{2n^2}$, we have $\sum\limits_{k=1}^n\frac{ku_k}{n^2}\le\epsilon\frac{n^2+n}{2n^2}$. – robjohn Dec 2 '17 at 19:22