This is a question from the book Methods of Real Analysis by R. R. Goldberg.

If $(s_n)$ is a sequence of real numbers and if $$\sigma_n=\frac{s_1+s_2+\cdots+s_n}{n}$$ then prove that: $\operatorname{{lim sup}}\sigma_n \leq \operatorname{lim sup} s_n$.

I don't have any idea how to start working on this problem. Please help. Thanks.

  • $\begingroup$ well, you may consider that $\sup \sigma_n \leq \sup \ s_n$ for all n $\endgroup$ – Mathematics Sep 9 '12 at 13:26
  • $\begingroup$ @Mathematics: How shall I prove that? And how will it help in this problem? Please explain. $\endgroup$ – Sayantan Sep 9 '12 at 13:33
  • $\begingroup$ Assume the opposite, that there is some $k$ with $\operatorname{lim sup} s_n \lt k \lt \operatorname{lim sup}\sigma_n$, and find a contradiction $\endgroup$ – Henry Sep 9 '12 at 13:37
  • $\begingroup$ ${\sigma_n = \frac{s_1+s_2+\cdots+s_n}{n}} \le \frac{n(\sup s_n)}{n}$ and take $\lim$ on both side. $\endgroup$ – Mathematics Sep 9 '12 at 15:22
  • $\begingroup$ This is a consequence of Stolz-Cesaro theorem, see e.g. here. $\endgroup$ – Martin Sleziak Sep 9 '12 at 20:40

Fix an integer $k$. Let $n\geqslant k$. Then $$\sigma_n=\frac 1n\sum_{j=1}^ks_j+\frac 1n\sum_{j=k+1}^ns_j\leqslant \frac 1n\sum_{j=1}^ks_j+\frac{n-k}n\sup_{l\geqslant k}s_l\leqslant \frac 1n\sum_{j=1}^ks_j+\sup_{l\geqslant k}s_l.$$ Now take on both sides the limsup when $\color{red}{n\to +\infty}$: we get the wanted result.

Taking $s_n:=(-1)^n$, we can see that the inequality may not be an equality.

  • $\begingroup$ Can you prove it like this: ${\sigma_n = \frac{s_1+s_2+\cdots+s_n}{n}} \le \frac{n(\sup s_n)}{n}$ and take lim on both sides ? This seems easier to me, but it seems to easy to be right. $\endgroup$ – Kasper Feb 7 '13 at 17:53
  • 2
    $\begingroup$ Where do you take the supremum? $\endgroup$ – Davide Giraudo Feb 7 '13 at 19:46
  • $\begingroup$ @Kasper This only proves the strictly weaker inequality $$\limsup_n\sigma_n\leqslant\sup_ns_n.$$ $\endgroup$ – Did Sep 27 '14 at 6:30
  • $\begingroup$ Another question asking for clarification for this answer was posted. (I tried to answer it, but in any case I thought it would be polite to leave you a ping. Maybe you will have something to say to the OP of the other question, too.) $\endgroup$ – Martin Sleziak Dec 3 '14 at 13:43
  • $\begingroup$ @MartinSleziak Thanks for pinging. $\endgroup$ – Davide Giraudo Dec 3 '14 at 13:47

Here's a simple solution:

Let $x^{*}_k = \sup_{k \geq n} \{x_n\}$. Then, $$x^{*}_k \to \limsup_{n \to \infty} \{x_n\}.$$

By a simple fact, a convergent sequence's averages converge to the same limit, so $$\sigma_n^{*} = \frac{1}{n} \sum_{j=1}^{n} x^{*}_j \to \limsup_{n \to \infty} \{x_n\}$$ as well. Now since $\sigma_n \leq \sigma^{*}_n$, the result follows.


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