Let $\{s_n\} \subset \mathbb{C}$. Define $$\sigma_n := \frac{s_0 + \cdots + s_n}{n+1}.$$ (e) Let $a_n := s_n - s_{n-1}$ for $n \geq 1$. Assume further that $|na_n| \leq M$ for all $n$, and $\sigma_n \to \sigma$. Prove that $\lim s_n = \sigma$. Hint: for $m < n$, we have that $$s_n - \sigma_n = \frac{m+1}{n-m}(\sigma_n - \sigma_m) + \frac{1}{n-m} \sum_{i=m+1}^n (s_n - s_i).$$ For $m+1 \leq i \leq n$, we have that $$ |s_n - s_i| \leq \frac{(n-i)M}{i+1} \leq \frac{(n-m-1)M}{m+2}. $$ Fix $\epsilon > 0$ and for each $n$, we find some $m$ such that $$m \leq \frac{n-\epsilon}{1+\epsilon} < m+1.$$ Then it follows that $\frac{m+1}{n-m} \leq \epsilon^{-1}$ and $|s_n - s_i| < M\epsilon$. Hence $\limsup_{n\to \infty} |s_n - \sigma| \leq M\epsilon$.

Following the hint, I think our goal is to show that $\limsup_{n \to \infty} |s_n - \sigma_n| = 0$. I could prove the first equality concerning $s_n - \sigma_n$, and get the second inequality regarding $|s_n-s_i|$. Then plug in, we have that $$ |s_n - \sigma_n| \leq \underbrace{\frac{m+1}{n-m}}_{(\text{I})}(\sigma_n - \sigma_m) + \underbrace{\frac{n-m-1}{m+2}}_{(\text{II})}M. $$ Since we have the convergence of $\sigma_n$, by Cauchyness, we could control $(\sigma_n - \sigma_m)$. Therefore at this stage, we would like to control (I) and (II), so that (I) is somehow "bounded", and make (II) as small as we want.

The hint suggests that we should find corresponding $m$, then if make $n, m$ large. so that $\sigma_n - \sigma_m \leq \epsilon^2$, then we are done.

My Question: Without the hint, I think it is natural to come up with the first equality and the second inequality. But how to come up with such $m$, to control (I) and (II)? More precisely, why would we choose the integer $m$ to satisfy that

$$m \leq \frac{n-\epsilon}{1+\epsilon} < m+1?$$

I am afraid that my description is somewhat vague, so please let me know if I could make it more clear. Thank you!


1 Answer 1


I find it useful to write down what you need and then work your way back to check if these conditions may be satisfied. In your case, as you said, you want to control (I) and (II) such that (I) is bounded and (II) may be arbitrarily small.

Then you may assume that (I) is bounded by $A>0$: $$\frac{m+1}{n-m}\leq A\iff m\leq\frac{An-1}{A+1}$$ and also assume that (II) is less than any arbitrarily small $\epsilon>0$: $$\frac{n-m-1}{m+2}\leq\epsilon\iff\frac{n-1-2\epsilon}{\epsilon+1}\leq m$$ Combining them you have $$\frac{n-1-2\epsilon}{\epsilon+1}\leq m\leq\frac{An-1}{A+1}\qquad(*)$$ Since you are trying to estimate $|s_n-\sigma_n|$, in which the index $m$ does not appear, you are free to choose any $m$ that satisfies $(*)$ (and also $m<n$, but this is implied by $(*)$).

So until now, the question you should be asking is: is there a constant $A>0$ such that for all sufficiently small $\epsilon$ and all sufficiently large $n$, the set $$\{m:m\text{ satisfies }(*)\}$$ is nonempty. A lucky guess says $A=\epsilon^{-1}$ is a possible choice of $A$, which is exactly what the hint gives.

As $\epsilon$ may be arbitrarily small, $A=\epsilon^{-1}$ can be arbitrarily large, hence we may not consider (I) to be bounded. But we can still make use of this estimate with some additional remarks. For any given $\epsilon>0$, we choose $N>0$ such that $$m,n>N\implies|\sigma_m-\sigma_n|<\epsilon^2$$ Then consider only $n>3N$. WLOG assume $N>4$ and $\epsilon<1/2$. Then if $m$ satisfies $(*)$ we have $$m\geq\frac{n-1-2\epsilon}{\epsilon+1}>\frac{n-2}{3/2}>n/3>N\\ \implies|\sigma_m-\sigma_n|<\epsilon^2\\ \implies\frac{m+1}{n-m}|\sigma_m-\sigma_n|\leq\frac{\epsilon^2}{\epsilon}=\epsilon$$ fixing the problem of (I) being possibly not bounded.

  • $\begingroup$ I agreed with what you said. But why (*) implies that $m < n$? Btw, I think there is a typo: the hint should suggest $A := \epsilon^{-1}$. $\endgroup$
    – mathdoge
    May 8, 2019 at 12:10
  • $\begingroup$ $(*)$ implies $m<n$ because $m\leq\frac{An-1}{A+1}<\frac{An}{A+1}<n$. Also, I have thought about the typo. It turns out I have to make some other changes as well (see my edited answer). The choice of $A$ may not be so free after all. My new idea is that, for $(*)$ to contain an integer $m$, it is sufficient that $\frac{n-1-2\epsilon}{\epsilon+1}+1\leq\frac{An-1}{A+1}$. And $A=\epsilon^{-1}$ is the smallest possible number for this to hold (since we are bounding (I) by $A$, we want $A$ to be as small as possible). $\endgroup$
    – trisct
    May 8, 2019 at 12:58
  • $\begingroup$ Thank you again for your clarification, I got your idea! Previously I was also concerned about the existence of $m$ as an integer. It turns out that $A \leq \epsilon^{-1}$, where the bound depends on $\epsilon$, but still controllable! $\endgroup$
    – mathdoge
    May 8, 2019 at 13:32

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