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We define:

series $\sum_{n=1}^{\infty} a_n$

$s_n=\sum_{k=1}^{n} a_k, n\in \mathbb{N}.$

$\sigma_n=\frac{1}{n}\sum_{k=1}^n s_k$

$s=\lim_{n\to \infty} s_n$

Proposition: If sequence $(s_n)_n$ converges and if $s=\lim_{n\to \infty} s_n$ then sequence $(\sigma_n)_n$ also converges and $s=\lim_{n\to \infty} \sigma_n$.

Proof: Let $s=\lim_{n\to \infty} s_n$ and $\epsilon \gt0$. Let $\epsilon_n=s_n-s,n\in \mathbb{N}$.There exists $m\in \mathbb{N}$ so that $\forall n\in \mathbb{N}, (n\gt m) \Rightarrow (|\epsilon_n \lt \frac{\epsilon}{3}|)$. Now for $n \gt m+1$ we have $$\sigma_n=\frac{s_1+...+s_{m+1}}{n}+\frac{s_{m+2}+...+s_n}{n} =$$ $$\frac{s_1+...+s_{m+1}}{n}+\frac{n-m-1}{n}s+\frac{\epsilon_{m+2}+...+\epsilon_n}{n}$$

From that we have (and I don't understand why)$$|\sigma_n-s|\lt \frac{|s_1+...+s_n|}{n}+\frac{m+1}{n}|s|+\frac{n-m-1}{n}\frac{\epsilon}{3}$$

Lets take $n_{\epsilon} \gt m+1$ so that $\frac{|s_1+...+s_n|}{n_{\epsilon}}\lt \frac{\epsilon}{3}$ and $\frac{(m+1)|s|}{n_{\epsilon}} \lt \frac{\epsilon}{3}$.

Now, $\forall n\in \mathbb{N}, (n \gt n_{\epsilon}) \Rightarrow |\sigma_n-s|\lt \epsilon$

Q.E.D

If someone could explain that transition, I'd be happy.

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Subtract $s$ from $\sigma_n$, take absolute values, and apply the triangle inequality:

$$\begin{align*} \left|\sigma_n-s\right|&=\left|\frac{s_1+\ldots+s_{m+1}}n-\frac{m+1}ns+\frac{\epsilon_{m+2}+\ldots+\epsilon_n}n\right|\\ &\le\frac{|s_1+\ldots+s_{m+1}|}n+\frac{m+1}n|s|+\frac{|\epsilon_{m+2}|+\ldots+|\epsilon_n|}n\\ &<\frac{|s_1+\ldots+s_{m+1}|}n+\frac{m+1}n|s|+\frac{n-m-1}n\cdot\frac{\epsilon}3\;. \end{align*}$$

The final step uses the fact that each $|\epsilon_k|<\frac{\epsilon}3$, and there are $n-m-1$ of these $\epsilon_k$ terms in the numerator.

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First, note that $$ \left|\frac{n-m-1}{n}s-s\right|=\left|\frac{n-m-1}{n}s-\frac{n}{n}s\right|=\left|\frac{-m-1}{n}s\right|=\frac{m+1}{n}\left|s\right|. $$ Next, since $m$ is chosen so that $|\epsilon_{n}|<\epsilon/3$ for all $n>m$, $$ \left|\frac{\epsilon_{m+2}+\cdots+\epsilon_{n}}{n}\right| \leq \frac{|\epsilon_{m+2}|+\cdots+|\epsilon_{n}|}{n} <\frac{\overbrace{\epsilon/3+\cdots+\epsilon/3}^{n-m-1\text{ terms}}}{n}=\frac{n-m-1}{n}\frac{\epsilon}{3}. $$

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  • $\begingroup$ If you found any of the answers helpful, please consider upvoting or accepting; this is a good policy to follow on all questions you ask. (Personally, I think Brian's answer is nicer than mine, so consider accepting that one ^_^) $\endgroup$ – parsiad Sep 4 '16 at 17:19
  • $\begingroup$ I think that it’s a tossup! $\endgroup$ – Brian M. Scott Sep 4 '16 at 17:39

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