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I have read this solution, but I could not understand it.

It has shown $$\sigma_n\leqslant \frac 1n\sum_{j=1}^ks_j+\sup_{l\geqslant k}s_l,$$ but how does it go to $$\sup(\sigma_n)\leqslant \frac 1n\sum_{j=1}^ks_j+\sup_{l\geqslant k}s_l$$

I have thought of regarding the RHS as upper bound of $\sigma_n$, but I am not sure about this, since RHS varies as $n$ changes.

The writer wrote "take on both sides the limsup when $n \to \infty$", but I could not understand it. Please help.

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k99731
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2 Answers2

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First I will copy the relevant part of the answer you are talking about so that we have some context:

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

I would slightly change the formulation and the conclusion. (I think we need to add one more step to finish the conclusion. But I might have missed something obvious. Or maybe Davide considered the missing step easy and left it for the reader.)

So we have the inequality $$\sigma_n \le \frac 1n\sum_{j=1}^ks_j+\sup_{l\ge k}s_l.$$ This equality is true for any $n\ge k$.

If we take limit superior w.r.t. $n$ on both sides we get $$\limsup_{n\to\infty} \sigma_n \le \sup_{l\ge k}s_l.$$ (Since the first term on the RHS converges to zero for $n\to\infty$ and the second one does not depend on $n$.)

Now the above inequality is true for arbitrary $k$. So we also have

$$\limsup_{n\to\infty} \sigma_n \le \lim_{k\to\infty} \sup_{l\ge k}s_l = \limsup_{k\to\infty} s_k.$$

Martin Sleziak
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  • This clears things up quite a bit. The "fixed $k$" might otherwise deflect one from thinking of that last step. I don't yet see a way to get the conclusion by taking lim sup just once, even if we let $k$ be a function of $n$. – David K Dec 03 '14 at 13:54
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    The last inequality in the above chain holds only if $\sup_{l\geqslant k}s_l$ is non-negative, as was pointed out in this request for clarification. But you can simply omit it and take the $\limsup$ in the last-but-one expression. – Martin R Nov 01 '19 at 18:30
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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.

Anthony Peter
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