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Prove that the divergence of the following sequence. $$s_n=\frac{(-1)^nn}{2n-1}$$

The following is the sample answer

Note that $\exists N\in\mathbb{N}\, s.t.\,\forall k\geq N$ $$|s_{2k}-1/2|<1/2$$ $$|s_{2k+1}+1/2|<1/2$$

So $$l\geq 0\implies|s_{2k+1}-l|>1/4$$ $$l<0\implies|s_{2k}-l|>1/4$$

$s_n$ diverges.

First I do not know where the above four inequalities come from. Second I do not know why the last two inequalities implies that $s_n$ diverges. For me, what I will try to prove alternating sequence diverge is suppose that $s_n$ will converge to $s$ and prove that $s$ does not exist to show contradiction. Or, I may prove that $\lim \sup(s_n)\neq \lim \inf(s_n)$

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  • $\begingroup$ Bad idea: $|s_{2k}-1/2|<1/2$ and $|s_{2k+1}+1/2|<1/2$ do not imply divergence. $\endgroup$ – Did Dec 27 '14 at 11:08
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Hint

You know that $(s_n)$ converge to $\ell$ if all subsequence converge to $\ell$. Now, consider the subsequences $(s_{2k})$ and $(s_{2k+1})$, what can you conclude ?

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  • $\begingroup$ I want to know how to establish the first two inequalities and what is $l$. $\endgroup$ – Alan Wang Dec 27 '14 at 11:17
  • $\begingroup$ You have that $$s_{2k}\longrightarrow \frac{1}{2}$$ and $$s_{2k+1}\longrightarrow -\frac{1}{2}$$ therefore $(s_n)$ doesn't converge ! $\endgroup$ – idm Dec 27 '14 at 11:33
  • $\begingroup$ This is a very good hint, but I think you're using that if $(s_n)$ converges to $\ell$, then all subsequences converge to $\ell$. $\endgroup$ – user84413 Dec 27 '14 at 21:28

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