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Looking for verification of a proof. Find $\{s_n\}$ a sequence of positive numbers such that $\limsup_{n\to\infty}s_n=\infty$, and $\lim_{n\to\infty}\sigma_n=0$ where $\sigma_n=\frac1{n+1}\sum_{k=0}^ns_n$.

Let $$s_n= \begin{cases} m+1, & \text{if $n=2^m$}\\ \frac1{2^n}, & \text{otherwise} \end{cases}$$

If $2^m\le n<2^{m+1}$,

$$\begin{align} \sigma_n & =\frac1{n+1}\left(\sum_{k=0}^n\frac1{2^k}+\sum_{k=0}^mk-\sum_{k=0}^m\frac1{2^{2^k}}\right)\\ & =\frac1{n+1}\left(2-\frac1{2^n}+{(m+2)(m+1)\over 2}-\sum_{k=0}^m\frac1{2^{2^k}} \right). \end{align}$$

Now for $2^m\le n<2^m-1$,

$$\begin{align} \sigma_n-\sigma_{n+1} & =\frac1{n+1}\left(2-\frac1{2^n}\right)-\frac1{n+2}\left(2-\frac1{2^{n+1}}\right)\\ & +\frac1{(n+1)(n+2)}\left({(m+2)(m+1)\over 2}-\sum_{k=0}^m\frac1{2^{2^k}}\right). \end{align}$$

Note that the last term is clearly positive, and

$$\begin{align} & \frac1{n+1}\left(2-\frac1{2^n}\right)-\frac1{n+2}\left(2-\frac1{2^{n+1}}\right)\\ = & {2\over (n+1)(n+2)}+{1\over 2^{n+1}(n+2)}-{1\over 2^n(n+1)}\\ = &{2^{n+2}-(n+3)\over 2^{n+1}(n+1)(n+2)}. \end{align}$$

This difference is also positive because $2^{n}>n+1$ for $n\ge2$. So, $$\max\{\sigma_n:2^m\le n\le 2^{m}-1\}=\sigma_{2^m}.$$

Therefore to prove $\lim_{n\to\infty}\sigma_n=0$, it suffices to prove $\lim_{m\to\infty}\sigma_{2^m}=0$.

$$\begin{align} \lim_{m\to\infty}\sigma_{2^n} & \le \lim_{m\to\infty}\frac1{2^m+1}\left(2-\frac1{2^{2^m}}+{(m+2)(m+1)\over 2}\right)\\ & =\lim_{m\to\infty}\frac2{2^m+1}-\lim_{m\to\infty}{1\over (2^m+1)2^{2^m}}+\lim_{m\to\infty}{(m+1)(m+2)\over 2(2^m+1)}\\ & = 0. \end{align}$$

Tim Raczkowski
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    I don't know if your solution is correct because it's huge and it makes my head hurt to look at it, so I haven't read it. I think I am not alone in this because this question has gone unanswered for quite awhile. There exist simpler solutions, and I'm pretty sure you can find one. – Matt Samuel Jan 15 '15 at 02:36
  • @MattSamuel Yes, I believe that you are right. – Tim Raczkowski Jan 15 '15 at 02:38

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Your example $s_n$ works, and the proof looks reasonable, although I haven't gone through every computation. However, I strongly suggest refactoring it as follows.

  1. Consider the sequence defined by $a_n = k$ when $n=2^k$ and $a_n=0$. It is unbounded. The partial sums $s_n=a_1+\dots+a_n$ are given by $k(k+1)/2$ where $k\le \log_2 n$. Thus, $s_n/n\to 0$.

  2. Consider the sequence $b_n=1/n$. It converges to zero, therefore its means also converge to zero: (standard fact, proved here).

  3. Conclude from 1 and 2 that the sequence $s_n=a_n+b_n$ has the desired properties.

Community
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  • Thanks for the input. I will go over you suggestions. – Tim Raczkowski Jan 15 '15 at 18:05
  • Yes, this is a much nicer way of solving the problem. As has been pointed out, the calculations of my solution were quite involved. I'm fairly confident that they are correct, but, as always, it better to work smart than to work hard. :) – Tim Raczkowski Jan 15 '15 at 19:28