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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 $ ... 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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    @Daniel, any thought on the solution below?2011-08-27

1 Answers 1

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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.