Provided that $\lim \limits_{n\to \infty} u_n=l$, prove that $\lim \limits_{n\to \infty} \frac{\sum_{k=1}^nu_k}{n}=l$

Question

The question is from Schaum's Outline of Complex numbers

This is how I have gone about it

$|\frac{\sum_{k=1}^nu_k}{n}-l|$

$=|\frac{\sum_{k=1}^n (u_k-l)}{n}|$

$\le \frac{1}{n}\sum_{k=1}^n|u_k-l|$

Now, since $\lim \limits_{n\to \infty} u_n=l$, $\lim \limits_{n\to \infty} (u_n-l)=0 $ $\to \lim \limits_{n\to \infty} |u_n-l|=0$

I know that this is a necessary condition for $\sum_{k=1}^n|u_k-l|$ to converge, but not a sufficient one. If it does converge, then we are done, since we can say that $\sum_{k=1}^n|u_k-l|=S$, where $S$ is finite, so $|\frac{\sum_{k=1}^nu_k}{n}-l| \le \frac{1}{n}\sum_{k=1}^n|u_k-l|=\frac{S}{n}$, so if $\frac{S}{n}<\epsilon$ i.e $n>\frac{S}{\epsilon}$, then $|\frac{\sum_{k=1}^nu_k}{n}-l|<\epsilon$

The problem is I don't know how to show that $\sum_{k=1}^n|u_k-l|$ actually converges.

Any help would be appreciated, thank you.

Apply Cesaro-Stolz, like here. Here is another similar question. — rtybase, May 06 '20 at 17:29
$ \left(\sum\limits_{k=1}^{n}{\left|u_{k}-\ell\right|}\right) $ may not converge, so don't try to show that. — CHAMSI, May 06 '20 at 17:45

Koro · Accepted Answer · 2020-05-06T18:14:24.410

Since, $\lim \limits_{n\to \infty} |u_n-l|=0$, we know that for every $\epsilon \gt 0$ there exists an N such that for all $n\gt N$ we have, $|u_n-l|\lt \epsilon$. Hence for $n \gt N$
$|\frac{\sum_{k=1}^nu_k}{n}-l|=|\frac{\sum_{k=1}^n (u_k-l)}{n}|\le \frac{1}{n}\sum_{k=1}^n|u_k-l|\le \frac{1}{n}[\sum_{k=1}^{N}|u_k-l|+\sum_{k=N+1}^{n}|u_k-l|]\\\lt\frac{1}{n}[\sum_{k=1}^{N}|u_k-l|+(n-N)\epsilon]$
Hence,
$\lim_{n\to \infty}|\frac{\sum_{k=1}^nu_k}{n}-l|\le\epsilon \tag{#}$
Since $\epsilon $ is any arbitrary positive no., we have $\lim_{n\to \infty}|\frac{\sum_{k=1}^nu_k}{n}-l|=0$, whence the result follows.

# I have used this result: if $(x_n), (y_n)$ are sequences such that $x_n \lt y_n$ for all $n\ge K\in N$, then $\lim_{n\to \infty} x_n\le \lim_{n\to \infty} y_n$ (of course, if the limits on both sides exist)

Thanks very much for this! You have answered my question. – Amrit Cassim May 06 '20 at 18:15 — Amrit Cassim, May 06 '20 at 18:15

score 1 · Answer 2 · answered May 06 '20 at 17:57

Let $ \varepsilon >0 $ :

Let $ n $ be a positive integer, we have : \begin{aligned}\left|\frac{1}{n}\sum_{k=1}^{n}{u_{k}}-\ell\right|&=\frac{1}{n}\left|\sum_{k=1}^{n}{\left(u_{k}-\ell\right)}\right|\\ &=\frac{1}{n}\left|\sum_{k=1}^{n_{0}-1}{\left(u_{k}-\ell\right)}+\sum_{k=n_{0}}^{n}{\left(u_{k}-\ell\right)}\right|\\ &\leq\frac{1}{n}\left|\sum_{k=1}^{n_{0}-1}{\left(u_{k}-\ell\right)}\right|+\frac{1}{n}\sum_{k=n_{0}}^{n}{\left|u_{k}-\ell\right|}\end{aligned}

Since $ u_{n}\underset{n\to +\infty}{\longrightarrow}\ell\in\mathbb{R} $, then there exists some $ n_{1}\in\mathbb{N} $, such that $ \left(\forall n\geq n_{1}\right),\ \left|u_{n}-\ell\right|<\varepsilon \cdot $

Since $ \left|\sum\limits_{k=1}^{n_{0}-1}{\left(u_{k}-\ell\right)}\right| $ does not depend of $ n $, $ \frac{1}{n}\left|\sum\limits_{k=1}^{n_{0}-1}{\left(u_{k}-\ell\right)}\right|\underset{n\to +\infty}{\longrightarrow}0 $, thus, there exists some $ n_{2}\in\mathbb{N} $, such that $ \left(\forall n\geq n_{2}\right),\ \frac{1}{n}\left|\sum\limits_{k=1}^{n_{0}-1}{\left(u_{k}-\ell\right)}\right|<\varepsilon \cdot $

Hence, if $ n $ is such that $ n\geq\max\left(n_{1},n_{2}\right) $, we have that : \begin{aligned}\left|\frac{1}{n}\sum_{k=1}^{n}{u_{k}}-\ell\right|\leq\varepsilon +\frac{1}{n}\sum_{k=n_{0}}^{n}{\varepsilon}=\left(2-\frac{n_{0}}{n}\right)\varepsilon<2\varepsilon\end{aligned}

Thus : $$ \frac{1}{n}\sum_{k=1}^{n}{u_{k}}\underset{n\to +\infty}{\longrightarrow}\ell $$

Provided that $\lim \limits_{n\to \infty} u_n=l$, prove that $\lim \limits_{n\to \infty} \frac{\sum_{k=1}^nu_k}{n}=l$

2 Answers2