Consider the series
$$\sum_{n=0}^{\infty}\cos^n(n)$$
I think that the root test is inconclusive, because
$$\limsup_n \sqrt[n]{|\cos^n(n)|}=\limsup_n|\cos(n)|\leq 1$$
once we can approximate $\pi$ by rational numbers, there will always be some $i$ and $j\in\mathbb{N}$ such that $|j\pi-i|<\varepsilon$, for every $\varepsilon>0$ that we choose. And in this case $|\cos(i)-1|<\delta$.
Nevertheless, it seems that it converges. I can't think of any other convergent series to compare with it.
My question is: how can I prove that this series converges?
Edit: Actually, this series diverges, as you can see in tmyklebu's answer. I made a fortran program and here are some values of the sequence of the partial sums:
n S_n
10 1.5898364866640549
100 7.8365722183614510
1000 24.825953005207236
10000 79.232008037801393
It doesn't. That series has a lot of terms near $\pm 1$; in particular, I can show it has infinitely many terms whose absolute value is larger than $\frac12$:
The inequality $|\pi - p/q| < 1/q^2$ is satisfied by infinitely many pairs of positive integers $(p,q)$. (This is Dirichlet's approximation theorem.) Let $(p,q)$ be one such pair with $q > 8$. Then, for some real $r$ with $|r| < 1/q$, we have $$|\cos(p)| = |\cos(q \pi + r)| \geq 1 - r^2 \geq 1 - 1/q^2.$$ Then $$|\cos(p)^p| \geq (1 - 1/q^2)^p \geq 1 - p/q^2 \geq 1 - 4/q > 1/2.$$