Find supremum and infimum of $\left\lbrace\frac{\cos n\pi}{n^2+1}\right\rbrace^{\infty}_{n=1}$

Question

How do I find supremum and infimum of $\left\lbrace\frac{\cos n\pi}{n^2+1}\right\rbrace^{\infty}_{n=1}$?

I know that the function $\left\lbrace\frac{\cos n\pi}{n^2+1}\right\rbrace$ is neither increasing nor decreasing as it behaves differently for even and odd n. So therefore we must employ n to $n=2k$ and $n=2k+1$ and then we have to solve both: $\left\lbrace\frac{\cos 2k\pi}{(2k)^2+1}\right\rbrace^{\infty}_{n=1}$ and $\left\lbrace\frac{\cos(2k+1)\pi}{(2k+1)^2+1}\right\rbrace^{\infty}_{n=1}$.

But I would appreciate help with how I am going to continue from here?

Thanks in advance!

Answer 1

Exactly employing even and odd sub-sequences gives answer. Using $\cos n\pi = (-1)^n$, one is $\left\lbrace\frac{1}{(2k)^2+1}\right\rbrace^{\infty}_{k=1}$. It's decreasing, with maximum first member and infimum = limit = $0$. And second one $\left\lbrace\frac{-1}{(2k-1)^2+1}\right\rbrace^{\infty}_{k=1}$, which is increasing with first member minimum and with supremum = limit = $0$.

So initial sequence have maximum, which is also supremum, and minimum, equals infimum, accordingly considered subsequences.

By the way, sequence have limit, which equals $0$.