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Let $(a_n)$ be a sequence defined as $$ a_n = (-1)^n \frac{3n + 5}{n + 1}. $$ I need to compute the $\lim \inf$ and $\lim \sup$ of this. Here is what I did. By definition we have \begin{align*} \lim_{n \to \infty} \sup a_n = \lim_{n \to \infty} \sup \left\{ (-1)^m \frac{3m + 5}{m + 1} \ | \ m \geq n \right\}. \end{align*} So $\lim \sup a_n = 3$ when $m$ is even, and $\lim \sup a_n = -3$ if $m$ is odd. For the $\lim \inf$ I think we have $\lim \inf a_n = \frac{3n + 5}{ n + 1} $ when $m$ is even, and $\lim \inf a_n = - \frac{(3n + 5)}{n + 1}$ when $m$ is odd.

Can someone correct me if I'm wrong please. I'm new to this concept and still trying to grasp it.

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  • $\begingroup$ $b_n = sup \{(-1)^m\frac{3m+5}{m+1} | m \ge 2n \} = \frac{6 n+5}{2 n+1}$ so $\lim_{n \to \infty} \sup (-1)^n\frac{3n+5}{n+1} = \lim_{n \to \infty} b_n = 3$ $\endgroup$ – reuns Jan 26 '16 at 16:01
  • $\begingroup$ If you take $a_{2n}$ and $a_{2n+1}$ it would be probably easier to find sup and inf. $\endgroup$ – user8469759 Jan 26 '16 at 16:02
  • $\begingroup$ This doesn't really help. Can you post a comprehensive answer? $\endgroup$ – Kamil Jan 26 '16 at 18:04

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