arithmetic mean of a sequence converges

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 prove 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 is $$<\frac{\varepsilon}2$$, 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 I have to take the first sum from $$k=1$$ to $$M$$ or can you do it as above?

Thanks for helping.

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".

• 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 $$\Bigl|\frac1n\sum_{k=1}^na_k-a\Bigr|\le M\,\frac{m}{n}+\frac1n\sum_{k=m+1}^n|a_k-a|\le\frac\epsilon2+\frac{n-m}{n}\frac\epsilon2\le\epsilon.$$

• Is the converse true? – kathystehl May 30 '15 at 23:24
• No. Example: $(-1)^n$. – Julián Aguirre May 31 '15 at 7:07
• the sequence is not bounded both-side – user284275 Feb 10 '17 at 18:24

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.

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