I'm doing a problem in which I should find the limit superior and inferior of the sequence $\left\{ \frac 1 n \cos{\frac{n\pi}{2}} \right\}^\infty_{n=1}$
If $n$ is even, then $\cos{\frac{n\pi}{2}}=(-1)^n$ so we have the two subsequences $s_1 = \left\{ -\frac 1 n \right\}$ and $s_2 =\left\{ \frac 1 n \right\}$.
Otherwise is n is odd, $\cos{\frac{n\pi}{2}}=0$, giving the subsequence $s_3=\left\{ 0 \right\}$.
Since these are the possible subsequences, and $s_1<s_3<s_2$, does this mean that $s_1$ and $s_2$ are the limits inferior and superior, respectively?
Use Cuchy’s theorem by starting as cos nπ/2 lies between -1 and 0.(been getting some technical problems with my pc, so can't type it mathematically) since when n is odd, cos nπ/2=-1 and when is even, cos nπ/2=0. Hence when we consider -1/n < (1/n)cos nπ/2 < 0. As n→∞, (-1/n)→0. Hence both superior limit and inferior limit = 0.