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We had a theorem that the means of a sequence also converges:

Let $(a_n)_{n\in\mathbb N}$ be a convergent sequence. Then $\displaystyle \overline a_n=\sum_{k=1}^n \frac{a_k}n$ also converges.

I've tried to proove it:

$|\overline a_n-a|=\frac1n|\sum_{k=1}^n(a_k-a)|\leq\sum_{k=1}^{M-1}|a_k-a|+\sum_{k=M}^n|a_k-a|$

The second sum ist $<\frac{\varepsilon}2$, because there is an $M\in\mathbb N$ so that $(a_n)_{n\in\mathbb N}$ converges. Now you can consider all $n\geq\max\{M,\frac2{\varepsilon}\sum_{k=1}^{M-1}|a_k-a|\}$ and so $|\overline a_n-a|<\varepsilon$.

But can you also say that there is a $K\in\mathbb N$ such that $\frac1n\sum_{k=1}^{M-1}|a_k-a|<\frac{\varepsilon}2$ for all $n\geq K$? And do HAVE TO take the first sum from $k=1$ to $M$ or can you do it as above?

Thanks for helping.

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Let $a_n := 1$. Then $a_n \to 1$, but $\sum_{k=1}^n \frac{1}{k}$ is not convergent. Is there a typo in your claim? – saz Nov 30 '12 at 16:39
@saz yes, it has to be $a_k$ not $a_n$ - sorry ;b – user32778 Nov 30 '12 at 16:43
up vote 6 down vote accepted

There are still some typos in your proof (and it isn't written down clearly, in my oppinion).

So your proof seems to use this idea: Let $\varepsilon>0$, then there exists $M \in \mathbb{N}$ such that for all $n \geq M: |a_n-a| \leq \frac{\varepsilon}{2}$. Thus

$$|\bar{a}_n-a| \leq \frac{1}{n} \underbrace{\sum_{k=1}^{M-1} |a_k-a|}_{=:N} + \frac{1}{n} \underbrace{\sum_{k=M}^n |a_k-a|}_{\leq \frac{1}{2} \varepsilon \cdot n}$$

Now let $n \geq \max \{M, \frac{2}{\varepsilon} \cdot N\}=:M'$, then $|\bar{a}_n-a|\leq \varepsilon$.

Whether you take the first sum from $k=1$ to $M$ or to $M-1$ doesn't matter. Important is that

  • the first sum is finite and does not depend on $n$
  • in the second sum there are only $|a_k-a|$ such that $k$ is bigger or equal than $M$.

You could also sum from $k=1$ to $M+j$ for some $j \geq 0$.

Moreover, choose $K := \frac{2}{\varepsilon} \cdot N$, then

$$\frac{1}{n} \underbrace{\sum_{k=1}^{M-1}|a_k-a|}_{N} \leq \frac{1}{K} \cdot N \leq \frac{\varepsilon}{2}$$

for all $n \geq K$, so the answer to your (first) question is "yes".

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Is the converse true? – kathystehl May 30 '15 at 22:24
@kathystehl By converse you mean that the convergence of the arithmetic mean implies the convergence of the sequence? This does not hold true, just consider $a_n = (-1)^n$. – saz May 31 '15 at 4:42
Yes, that's what I meant. Thank you! – kathystehl May 31 '15 at 15:43
The second sum doesn't have n elements – Whyka Jul 7 '15 at 22:45
@Whyka No, but it has at most $n$ elements. – saz Jul 8 '15 at 5:23

Since $\{a_n\}$ converges, it is bounded, and so is $\{a_n-a\}$. Choose $M$ such that $$ |a_n-a|\le M\quad\forall n\in\mathbb{N}. $$ If $1\le m<n$, then $$ \Bigl|\frac1n\sum_{k=1}^na_k-a\Bigr|\le\frac1n\sum_{k=1}^m|a_k-a|+\frac1n\sum_{k=m+1}^n|a_k-a|\le M\,\frac{m}{n}+\frac1n\sum_{k=m+1}^n|a_k-a|. $$ Let $\epsilon>0$ be given. First choose $m$ such that $|a_n-a|\le\epsilon/2$ for all $n>m$. Once $m$ is chosen, choose $n_0>m$ such that $M\,m/n_0\le\epsilon/2$. Observe that $n_0$ depends $m$ that depends on $\epsilon$, so hat $n_0$ depends on $\epsilon$. Then, if $n\ge n_0$ we have $$ \Bigl|\frac1n\sum_{k=1}^na_k-a\Bigr|\le M\,\frac{m}{n}+\frac1n\sum_{k=m+1}^n|a_k-a|\le\frac\epsilon2+\frac{n-m}{n}\frac\epsilon2\le\epsilon. $$

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Is the converse true? – kathystehl May 30 '15 at 23:24
No. Example: $(-1)^n$. – Julián Aguirre May 31 '15 at 7:07

This problem is a direct application of Dirichlet's test for series convergence. Note that, since the sequence $(a_n)$ converges, then the sequence of partial sums is bounded.

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Wouldn't that imply that since $1/n\to 0$, it also holds that $\sum_{n=1}^M 1/n$ is bounded? – ItsmeMimi Oct 15 '15 at 4:56

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