Test for Divergence Using Divergence Criterion for Functional Limits

Question

Can I prove divergence of $\sum_{n=1}^\infty\frac{1}{\cos(n)+2}$ by showing that $\lim_{x\to \infty}\frac{1}{\cos(n)+2}$ does not exist? I know there is very simple way to solve it, but I am thinking about using $2k\pi$ and $(2k+1)\pi$ to substitute n, and since $\lim_{x\to \infty}\frac1{\cos(2k\pi)+2}$ and $\lim_{x\to \infty}\frac{1}{\cos((2k+1)\pi)+2}$ are not equal, the limit does not exist, so the series $\sum_{n=1}^\infty\frac{1}{\cos(n)+2}$ is divergent. My teacher told me that the choice of n must be integers, but I think it is unnecessary during the proof of the limit's existence. What's the problem with my proof?

Answer 1

Your teacher is right: the sum only involves the values at integer points, so even if the limit doesn't exist for real values, it could still exist and be zero for integer values. For example, if $f(x)=2^{-x}+\sin(\pi x)$ then $\lim_{x\to\infty}f(x)$ does not exist for real $x$, but the limit of $f(1),f(2),\ldots$ does exist and is $0$ (and $\sum_{n=0}^{\infty}f(n)=2$).

However, you can still make this work. $2k\pi$ is not an integer, so just take the closest integer to it, and call that $n(k)$. Now $n(k)$ is within $0.5$ of a multiple of $2\pi$, so $\cos n(k)\geq \cos 0.5>0.8$. This allows you to construct two different subsequences of integers which you can show don't tend to the same limit. (In fact, neither of these subsequences actually have a limit, but that's harder to prove and not important for your purposes.)