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Let $a_n$ be a bounded sequence such that $\left|a_n\right| < M$ for all $n$. Let $T_k$ be an integer sequence such that $\lim_{k\to\infty} T_k = \infty$ but $\lim_k \frac{T_k}{T_{k+1}} = 1$. Show that

$$\lim_k \frac{a_1+a_2+\cdots a_{T_k}}{T_k}=L\implies \lim_n \frac{a_1 + \cdots + a_n}{n}=L$$

This reminds me of Cesaro mean, but not quite, since $a_n$ are not necessarily convergent. I think I need to use $\lim \frac{T_k}{T_{k+1}} = 1$, but not sure how.

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closed as off-topic by Did, Claude Leibovici, Juniven, Namaste, Mostafa Ayaz Feb 5 '18 at 19:10

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Yes, this is certainly false without the assumption $\lim T_k/T_{k+1}=1$.

Hint: Let's write $$\sigma_n=\frac{a_1+\dots+a_n}{n}$$and

$$t_k=\sigma_{T_k}.$$

Suppose $T_k\le n<T_{k+1}$. Then $$\sigma_n=\frac{T_k}{n}t_k+\frac{a_{{}_{T_k+1}}+\dots+a_{n}}{n}=I+II.$$

Note that $$|II|\le\frac{n-T_{k}}{n}M\le\frac{T_{k+1}-T_k}{T_k}M,$$ so $$|\sigma_n-t_k|\le\left(\frac{T_k}n-1\right)|t_k|+\frac{T_{k+1}-T_k}{T_k}M.$$

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  • $\begingroup$ What is $s$ though? Did you mean $\sigma$? $\endgroup$ – Yuki Kawabata Feb 10 '18 at 0:49
  • $\begingroup$ typo........... $\endgroup$ – David C. Ullrich Feb 10 '18 at 1:13

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