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If $\{s_n\}$ is a complex sequence, define its arithmetic mean $\sigma_n$, by

$$\sigma_n=\frac{s_0+s_1+\cdots+s_n}{n+1}.$$

Put $a_n=s_n-s_{n-1}$ for $n\ge 1$. Assume $M<\infty$, $|na_n|\le M$ for all $n$, and $\lim\sigma_n=\sigma$. Prove that $\lim s_n=\sigma$, by completing to following outline:

If $m<n$, then $$s_n-\sigma_n={m+1\over n-m}(\sigma_n-\sigma_m)+\frac1{n-m}\sum_{i=m+1}^n(s_n-s_i).\tag{1}\label{1}$$

For the $k$ in the last term, show

$$|s_n-s_k|\le{n-m-1\over m+2}M.\tag{2}\label{2}$$

Fix $\epsilon>0$ and associate with each $n$ the integer $m$ that satisfies

$$m\le{n-\epsilon\over 1+\epsilon}<m+1.$$

Then $$(m+1)/(n-m)\le\frac1\epsilon \text{ and } |s_n-s_k|\le M\epsilon.\tag{3}\label{3}$$ Hence $$\limsup_{n\to\infty}|s_n-\sigma|\le M\epsilon.\tag{4}\label{4}$$

I've worked through $\eqref{1}-\eqref{3}$, but having trouble with $\eqref{4}$.

My attempt:

Choose $m$ and $n$ so that $|\sigma-\sigma_m|<\epsilon$, then

$$\begin{align} \limsup_{n\to\infty}|s_n-\sigma| & \le\limsup_{n\to\infty}|s_n-\sigma_n|+\limsup_{n\to\infty}|\sigma_n-\sigma|\\ & =\limsup_{n\to\infty}|s_n-\sigma_n|\\ & \le{m+1\over n-m}\limsup_{n\to\infty}|(\sigma_n-\sigma_m)|\\ &\quad\quad+\frac1{n-m}\limsup_{n\to\infty}\sum_{k=m+1}^n|s_n-s_k|\\ & \le\frac1\epsilon|\sigma-\sigma_m|+M\epsilon\\ &\color{red}{\le 1+M\epsilon}. \end{align}$$

I don't know how to do this without getting the $1$ term.

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I agree with most of your work. A couple of notes:

  • "Choose $m$ and $n$ so that $|\sigma-\sigma_m|<\epsilon$" doesn't really make sense since $m$ is predetermined by your choice of $n$ and $\epsilon$.
  • All of the instances of $n$ and $m$ should still be contained within the $\limsup$'s.

So with these small alterations you can still get $$\limsup_{n\to\infty}|s_n-\sigma|\leq\frac{1}{\epsilon}\limsup_{n\to\infty}|\sigma_n-\sigma_m|+M\epsilon.$$

But since $\{\sigma_n\}$ is convergent and $\frac{n-\epsilon}{1+\epsilon}<m+1$ implies $m\to\infty$ as $n\to\infty$, we have

$$\limsup_{n\to\infty}|\sigma_n-\sigma_m|=0$$

and the desired result follows.

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  • $\begingroup$ I see your point about all instances of $m$ and $n$ being inside $\limsup$. If I'm understanding correcg $\endgroup$ – Tim Raczkowski Jan 17 '15 at 8:44
  • $\begingroup$ Correctly, the point of the bound $(m+1)/(n-m)<1/\epsilon$ is to replace $(m+1)/(n-m)$ with a constant so that the first term can vanish as $n,m\to\infty$. The fact that $1/\epsilon$ gets large as $\epsilon$ gets small is irrelevant as long as we let $n,m\to\infty$ first. $\endgroup$ – Tim Raczkowski Jan 17 '15 at 9:01
  • $\begingroup$ Right, since $\epsilon$ is fixed in advance, we've shown that the term goes to zero no matter how small $\epsilon$ is. $\endgroup$ – AMPerrine Jan 17 '15 at 15:01

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