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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 .

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  • $\begingroup$ I can't understand how this would work ? could you please elaborate a bit more ? $\endgroup$ – Anis Souames Dec 2 '17 at 18:16
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    $\begingroup$ 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 $n\to \infty$. $\endgroup$ – Vinícius Novelli Dec 2 '17 at 18:19
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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 $$ \lim_{n\to\infty}\left|\,\sum_{k=1}^n\frac{ku_k}{n^2}\,\right|=0\tag5 $$

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  • $\begingroup$ $\frac {n^2+n}{2n^2} $ is greater than $\frac {1}{2} $, but otherwise, this seems fine. Just replace $\frac {\epsilon}{2} $ with $\epsilon $. $\endgroup$ – Michael L. Dec 2 '17 at 18:50
  • $\begingroup$ The limit is $\frac12$, so we don't need to replace $\frac\epsilon2$. $\endgroup$ – robjohn Dec 2 '17 at 19:06
  • $\begingroup$ So $M$ in (2) is the greatest term $u_n$ for which $n<=N$ ? $\endgroup$ – Anis Souames Dec 2 '17 at 19:08
  • $\begingroup$ This is the maximum of $|u_n|$ for all $n\le N$. $\endgroup$ – robjohn Dec 2 '17 at 19:11
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    $\begingroup$ 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 $\sum\limits_{k=1}^n\frac{ku_k}{n^2}\le\epsilon\frac{n^2+n}{2n^2}$. $\endgroup$ – robjohn Dec 2 '17 at 19:22

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