If $c_{n}=\frac{n\cos(n\pi)}{n+1}$, find $\limsup c_{n}$ and $\liminf c_{n}$.

Question

If $c_{n}=\frac{n\cos(n\pi)}{n+1}$, find $\limsup c_{n}$ and $\liminf c_{n}$.

Here is what I did:

$\left \{c_{n} \right \}=\left \{-\frac{1}{2},\frac{2}{3},-\frac{3}{4},\frac{4}{5},-\frac{5}{6},... \right \}=\left \{-\frac{1}{2},-\frac{3}{4},-\frac{5}{6},... \right \}\cup \left \{\frac{2}{3},\frac{4}{5},... \right \}$

Then $\limsup c_{n}=1$ and $\liminf c_{n}=-1$. Is this the correct answer and approach?

Answer 1

Yeah, the core idea and the results derived are correct.

You're essentially choosing the subsequences where, for $n\in\Bbb N$, $\cos(n\pi)$ achieves its maximum ($1$) for the limit supremum, and those where it achieves its minimum $(-1)$ for the limit infimum. After all, the limit supremum is the maximal limit for any of the subsequences, and limit infimum the minimum, so it makes sense to try to maximize/minimize oscillatory values where you can.

Then from there, the limit as $n\to\infty$ of $n/(n+1)$ is $1$, and the product of the two limits for the corresponding subsequences give you the desired limits.