# limit superior and inferior of a series

I was trying to make a problem, but I realized that the hypotheses are wrong, as for example, $a_n =-n!$ Satisfies the hypothesis, but the alternating sum is clearly bounded. Suppose that the hypothesis was that if it were limited, in this case, as it could prove that the remainder of the lim sup and lim inf, given that value?

I suppose I must calculate separately each cleaning, the problem is I do not know very well as you would

I realize that the sum of the pair is growing, and the odd is decreasing, but among them are not if they can be compared. But for example if $a_2$ is positive then have the following $$... <A_4 <a_2 <a_1 <a_ 3 <....$$ With this as I could go? I do not understand yet how to calculate upper and lower limits, anyone know where I can learn T_T? Thanks!

The problem is : Let $a_n$ a non increasing sequence , let $$s_n = \sum\limits_{k = 1}^n {a_k \left( { - 1} \right)^{k + 1} }$$ prove that $s_n$ it´s bounded ( false, by the counterexample, but assume that it´s bounded) and also prove that $$\lim \sup s_n - \lim \inf s_n = \lim a_n$$

Reading the end I realized that $a_n$ is bounded (because the last thing they ask me to prove assume that a_n limit exists)

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What is the problem you were trying to make? What specifically are you asking here? –  JavaMan Aug 21 '11 at 23:07
I´m not talking of alternating series test, and nothing, I´m only saying that these are the hypothesis. Assume that $a_n$ it´s bounded below, and nothing else. –  Daniel Aug 21 '11 at 23:32
Please try and state the problem first and then discuss it after. I also don't see any point in trying to prove something, if you have a counterexample to it. If you try to prove something which stands as false, and succeed, then have false theorems which in other words means you have an unsound logic. The wording here also isn't exactly clear... as this phrase "With this as I could go?" indicates. –  Doug Spoonwood Aug 21 '11 at 23:41
To me, the post starts at The problem is and ends at $=\lim a_n$. The paragraphs before this or after that do not exist. Then, and once one realizes a hypothesis is missing (which is not that $a_n\to0$...), this is a sound problem. –  Did Aug 21 '11 at 23:47
@Didier: yes. That would be the way of reading it. Thanks for the instructions. The problem is that I read the non-existent paragraphs and lost patience. Your interpretation is certainly the correct one. I removed my misleading comment. –  t.b. Aug 21 '11 at 23:56
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## 1 Answer

The missing hypothesis is that $(a_n)$ is nonnegative. With this hypothesis, as steps towards the solution, you could try to prove the intermediary facts below:

• For every $n\ge1$, $a_1-a_2\le s_{2n}\le s_{2n+1}\le a_1$.

• The sequence $(s_{2n})$ is nondecreasing and the sequence $(s_{2n+1})$ is nonincreasing.

• $\limsup s_n=\lim s_{2n+1}$ and $\liminf s_n=\lim s_{2n}$.

• $\lim s_{2n+1}-s_{2n}=\lim a_{2n+1}$ and $\lim a_{2n+1}=\lim a_{n}$.

Hope this helps. If one step is unclear to you, just tell me.

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