If $\sum a_n$ is a convergent series with $S = \lim s_n$, where $s_n$ is the nth partial sum, then $\lim_{n \to \infty} \frac{s_1+…+s_n}{n} = S$ [duplicate]

Let $\sum a_n$ be a convergent series, and let $S = \lim s_n$, where $s_n$ is the nth partial sum.

I need to prove the following:

$\lim_{n \to \infty} \frac{s_1+...+s_n}{n} = S$

How do I go about proving that proof?

Definition of a limit

$\lim_{n \to \infty} f(x) = L$ if for every number $\epsilon>0$ there is some number $\sigma >0$ such that $|f(x)-L| <\epsilon$ whenever $0<|x-a|<\sigma$

marked as duplicate by Martin Sleziak, Claude Leibovici, JonMark Perry, zhoraster, NamasteJan 20 '17 at 13:04

• What are your thoughts on this? Start with the definition of limit. – астон вілла олоф мэллбэрг Jan 18 '17 at 23:53
• So as n approaches infinity I need to show that it is equal to S – jeb650 Jan 18 '17 at 23:56
• Of course. I suggest you put up the definition of limit that you know, in your question, so that we are on the same wavelength. Once you know the definition of limit, you will know the starting point. I am trying to take you slowly through this, so that you understand and can solve similar questions later in this way. – астон вілла олоф мэллбэрг Jan 19 '17 at 0:02
• just edited my definition of a limit into the question – jeb650 Jan 19 '17 at 0:15
• Observe the below answer. It starts with taking $\epsilon>0$, and then uses the fact that $s_n \to S$ to extract an $N$. Some playing around then gives the answer. Don't look at the playing around, just the use of the limit definition should be observed in the below answer. That is your lesson: first, proceed from definition. – астон вілла олоф мэллбэрг Jan 19 '17 at 0:20

Let $\epsilon>0.$ Since $s_n\rightarrow S$ we can choose an $N$ such that $s_n>S-\epsilon$ for all $n>N.$ Let $n>N$. Then can write $$\frac{s_1 + \ldots + s_n}{n} = \frac{s_1 + \ldots + s_N}{n} +\frac{s_{N+1}+\ldots s_n}{n} > \frac{s_1 + \ldots + s_N}{n} + \frac{n-N}{n}(S-\epsilon).$$ Taking $\liminf$ of both sides gives $$\liminf_{n\rightarrow\infty}\frac{s_1 + \ldots + s_n}{n} \ge (S-\epsilon).$$ This is true for all $\epsilon>0,$ so we have $$\liminf_{n\rightarrow\infty}\frac{s_1 + \ldots + s_n}{n} \ge S.$$

By the same reasoning, can show that $$\limsup_{n\rightarrow\infty}\frac{s_1 + \ldots + s_n}{n} \le (S+\epsilon)$$ for any $\epsilon >0.$ so that $$\limsup_{n\rightarrow\infty}\frac{s_1 + \ldots + s_n}{n} \le S.$$ Since $\liminf \le \limsup,$ these inequalities yield $\liminf = \limsup = S,$ so the limit exists and it is equal to $S$.

• so does this show that the limit is S? – jeb650 Jan 19 '17 at 0:09
• @jeb650 I think so. What do you think? – spaceisdarkgreen Jan 19 '17 at 0:10
• just making my way through your answer and trying to understand it – jeb650 Jan 19 '17 at 0:10
• @jeb650 If it helps, remember we're fixing $N.$ The fact that it's $\ge S-\epsilon$ for any $\epsilon >0$ means that it's $\ge S.$ – spaceisdarkgreen Jan 19 '17 at 0:13
• I think that last edit makes thinks more clear, thanks, that looks like a correct solution to my question – jeb650 Jan 19 '17 at 0:16

The point is that:

$$\big\lvert\frac{s_1+s_2+\ldots+s_n}{n}-S\big\rvert=\big\lvert\frac{s_1-S+s_2-S+\ldots+s_n-S}{n}\big\rvert\le\frac{1}{n}(\lvert s_1-S\rvert+\lvert s_2\rvert-S+\ldots+\lvert s_n-S\rvert)=\frac{1}{n}\sum_{i=1}^N\lvert s_i-S\rvert+\frac{1}{n}\sum_{i=N+1}^n\lvert s_i-S\rvert$$

The first term goes to zero when $n\rightarrow\infty$

By choosing $N$ large enough one can bound each of $\frac{1}{n}\sum_{i=1}^N\lvert s_i-S\rvert$ and $\frac{1}{n}\sum_{i=N+1}^n\lvert s_i-S\rvert$ by $\frac{\epsilon}{2}$

$$\left|\frac{1}{n}\sum_{k=1}^n s_k - \sum_{k=1}^n a_k\right| =\left|\sum_{k=1}^n (\frac{n-k+1}{n} a_k - a_k) \right|$$ $$\leq \varepsilon \left| \sum_{k=1}^m a_k\right| + \left| \sum_{k=m+1}^n a_k\right|\leq \varepsilon (|S|+\delta) + \varepsilon'$$ where for any given $\varepsilon,\varepsilon',\delta >0\$, $m$ can be chosen so that the above holds for all sufficiently large $n$. Namely: First choose $m$ large enough so that $\left|\sum_{k=1}^m a_k - S\right| < \delta$ and $\left| \sum_{k=m+1}^n a_k\right|<\varepsilon'$. Next, choose $n_0$ large enough so that for all $n\geq n_0$ it is true that $|\frac{1-k}{n}| < \varepsilon$ for $k\leq m$. Thus, the difference of the two series becomes arbitrarily small; since one of them converges to $S$, the other one also converges to $S$.

You re-prove Cesaro's theorem. By definition there exists $N\in \mathbf N$ such that $$\lvert S_n-S\rvert <\varepsilon\quad\forall n>N.$$ We deduce that, for $n>N$, \begin{align} \biggl\lvert\frac{S_1+\dots+S_n}n-S\,\biggr\rvert&=\biggl\lvert\frac{S_1-S+\dots+S_N-S}n+\frac{S_{N+1}-S+S_n-S}n\,\biggr\rvert\\ &\le\biggl\lvert\frac{S_1-S+\dots+S_N-S}n\biggr\rvert+\biggl\lvert \frac{S_{N+1}-S}n\biggr\rvert+\dots+\biggl\lvert \frac{S_n-S}n\,\biggr\rvert\\ &\le\biggl\lvert\frac{S_1+\dots+S_N}n-\frac{NS}n\biggr\rvert+\biggl\lvert \frac{S_{N+1}-S}n\biggr\rvert+\dots+\biggl\lvert \frac{S_n-S}n\,\biggr\rvert\\ &\le\frac{\lvert S_1+\dots+S_N+NS\rvert}n+\frac{\lvert S_{N+1}-S\rvert}n+\dots+\frac{\lvert S_n-S\rvert}n\\ &\le\frac{\lvert S_1+\dots+S_N+NS\rvert}n+\frac{\varepsilon}n+\dots+\frac{\varepsilon}n =\frac{\lvert S_1+\dots+S_N+NS\rvert}n+\frac{(N-n)\varepsilon}n\\ &\le \frac{\lvert S_1+\dots+S_N+NS\rvert}n+\varepsilon\le 2\varepsilon \end{align} if $n$ is large enough.

Using the Stolz-Cesaro Theorem, we have

$$\lim_{n\to \infty}\frac{\sum_{k=1}^n\left(\sum_{j=1}^ka_j\right)}{n}=\lim_{n\to \infty}\sum_{j=1}^{n+1}a_j=S$$

If one is unfamiliar with Stolz-Cesaro, then on can proceed as follows. Let $\epsilon>0$ be given. Then, choose and fix $N$ so large that $|s_k-S|<\epsilon/2$ whenever $k>N$..

Thus, we can write

\begin{align} \left|\frac{\sum_{k=1}^n s_k}{n}-S\right|&=\left|\frac1n \sum_{k=1}^n (s_k-S)\right|\\\\ &\le \frac1n \sum_{k=1}^N |s_k-S|+\frac1n \sum_{k=N+1}^n |s_k-S|\\\\ &\le \frac1n \sum_{k=1}^N |s_k-S| +\left(1-\frac{N}{n}\right)\frac{\epsilon}{2}\\\\ &\le \frac1n \sum_{k=1}^N |s_k-S| +\frac{\epsilon}{2}\\\\ &\le \frac{\epsilon}{2}+\frac{\epsilon}{2}\\\\ &=\epsilon \end{align}

whenever $n>\max\left(N+1,\frac{2\sum_{k=1}^N |s_k-S|}{\epsilon}\right)$

And we are done!

• thanks for your answer, but we have not touched on this theorem before. Is there another explanation? – jeb650 Jan 18 '17 at 23:59
• Well, we can write $\sum_{k=1}^n s_k=\sum_{k=1}^N s_k+\sum_{k=N+1}^n s_k$. Then, given $\epsilon>0$, we choose and fix $N$ so that $S-\epsilon<s_k<S+\epsilon$. Can you finish now? – Mark Viola Jan 19 '17 at 0:08
• I've added another way forward. Please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark – Mark Viola Jan 22 '17 at 14:50