real analysis - How to prove that $sumlimits_{k= 0}^n frac{ku_k}{n^2}$ 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 .

Answer

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
$$\lim_{n\to\infty}\left|\,\sum_{k=1}^n\frac{ku_k}{n^2}\,\right|=0\tag5$$