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Let $\{a_{k}\}$ be a non-negative, increasing sequence. Consider

$$S_{n} = \frac{1}{n}\sum_{k=1}^{n}a_{k}.$$

I would like to show that if $$\lim_{n\rightarrow\infty}S_{n} = L,$$ then $$\lim_{k\rightarrow\infty}a_{k} = L.$$

I don't really have any clue how I could solve this problem. By definition, there exists $N$ such that

$$\lvert S_{n} -L\rvert <\epsilon$$

or

$$\frac{1}{n}\left\lvert \sum_{k=1}^{n}a_{k} - nL\right\rvert< \epsilon$$

for all $\epsilon>0$ whenever $n>N$ and

$$\frac{1}{n}\left\lvert \sum_{k=1}^{n}a_{k} - nL\right\rvert \leq \frac{1}{n}\sum_{k=1}^{n}\lvert a_{k}-L \rvert.$$

Now, if we define a new sequence $\{\alpha_{k}\}$ by

$$\alpha_{k} := \lvert a_{k} -L\rvert,$$

then I am pretty sure there is a theorem from harmonic analysis that states something like $\sum_{k=1}^{n}{\alpha_{k}}$ converges if and only if $(1/n)\sum_{k=1}^{n}{\alpha_{k}}$ converges too.

From here, by the Divergence Theorem, I would get that necessarily $\alpha_{k} \rightarrow L$ for $\sum{\alpha_{k}}$ to converge and hence $(1/n)\sum \alpha_{k}$ would also require $\alpha_{k}\rightarrow L$ to converge.

However, I can neither find a reference to this theorem I only vaguely remember and this still doesn't get me anywhere because I don't think the convergence of $(1/n)\sum \alpha_{k}$ is helpful anyway.

I have also tried proving this by contradiction too, but I would really prefer a direct proof.

JessicaK
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  • Certainly $n a_n \ge S_n$, so $\limsup a_n\ge L$. If $\limsup a_n>L+\epsilon$ then there is an $m$ such that $a_n \ge L+\epsilon$, and then $S_n \ge \sum_1^m a_i + (n-m)(L+\epsilon)$, for $n>m$. – kimchi lover Aug 04 '17 at 00:53
  • The reversed question may be found here. That is, given$\lim a_n=L$, prove that $\lim S_n=L$. – Simply Beautiful Art Aug 04 '17 at 01:42
  • This is easy consequence of Cesaro-Stolz. Since $a_{n} $ increases it either tends to a limit or to $\infty$. By Cesaro-Stolz $S_n$ does the same. It follows that $a_{n} \to L$ and further we don't need $a_{n}$ to be non-negative. – Paramanand Singh Aug 05 '17 at 09:32

3 Answers3

1

Since the sequence is increasing $a_n \rightarrow \sup \{a_k : k \in \mathbb {N}\}$

Let prove that $L = \sup \{a_k : k \in \mathbb {N}\}$. First for every $n$, $a_n \le L$. Indeed if there is $N$ that $a_N > L+\epsilon$ for $\epsilon > 0$. Since the sequence is increasing then for every $n \ge N$, $a_n > L+\epsilon$. Since the sequence is non negative, then $S_n > \frac{n - N}{n} (L+\epsilon)$ after tending $n$ to $\infty$ we have $L \ge L + \epsilon$ which is absurde.

On the other hand if for all $n$, $a_n < L - \epsilon$ for $\epsilon > 0$ then $S_n < L-\epsilon$. Tending $n$ to infinity $L \le L- \epsilon$ which is also absurde.

Kroki
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Intuitively, since $a_n$ is non-negative and monotone, there exists $f(x)$ that is non-negative, monotone, and $f:\mathbb R^+\mapsto\mathbb R^+$ such that $f(n)=a_n$, and we may have$$\frac1n\sum_{k=1}^na_k\sim\frac1n\int_1^nf(x)~\mathrm dx$$And then by L'Hospital's rule, we find that$$\lim_{n\to\infty}\frac{\int_1^nf(x)~\mathrm dx}n=\lim_{n\to\infty}f(n)=\lim_{n\to\infty}a_n$$To prove the integral limit equals the given limit, we may use the squeeze theorem:$$\frac1n\int_1^nf(x)~\mathrm dx\le\frac1n\sum_{k=1}^na_k\le\frac1n\int_1^{n+1}f(x)~\mathrm dx$$which follows from $f(x)$ being monotone.

Simply Beautiful Art
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If $a_n$ tends to infinity them $L$ is infinity and if $a_n$ tends to a finite limit $a$ then $L=a$.
This is standard and the proof is elementary. The result follows immediately by considering these two cases.

user376343
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Kavi Rama Murthy
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  • While your answer is correct, you need to elaborate a bit and perhaps mention the use of Cesaro-Stolz Theorem. Since the sequence increases the only two cases are finite limit and infinite limit. The lack of elaboration perhaps led to a downvote. I normally don't like any downvotes with correct answers, so here goes my +1. – Paramanand Singh Aug 05 '17 at 09:29
  • @ParamanandSingh Following your comment, I took a look at the Cesaro-Stolz theorem and after some thought, I do not see how to make it work in this case. Can you tell me what $a_n$ and $b_n$ you need to choose by using Wikipedia's notation for the statement of the theorem? Why is it that limit of another sequence depending on $a_n$ should follow immediately just from the knowledge that $a_n$ converges to $L$? That does not seem obvious to me at all. – JessicaK Aug 09 '17 at 04:49
  • @JessicaK: Take $A_{n} = nS_{n}, B_{n} =n$ so that $(A_{n+1}-A_{n})/(B_{n+1}-B_{n})=a_{n+1}$ and $A_{n} / B_{n} =S_{n} $. Now $a_{n}$ is increasing so it tends to a limit or to $\infty$ and by Cesaro-Stolz $S_{n} $ also does the same. Hence it is clear that $a_{n} \to L$. – Paramanand Singh Aug 09 '17 at 08:22
  • @ParamanandSingh Thank you! I completely missed the idea of using $nS_{n}$ and nothing was coming out quite right. – JessicaK Aug 09 '17 at 10:02