$$ \sum_0^\infty \cos n\theta $$
The answer in the book says that this series is divergent. Which I initially agreed with because according to one of the theorems If $a_n = \cos n\theta$ and the sequence does not converge to $0$ then the series does not converge.
But then if the $\cos\theta$ graph is always moving in between $1$ and $-1$ shouldn't the summation equal $0$?
If the series converges, we must have $\lim_{n\to\infty}\cos n\theta=0$. In particular, there must be an $N$ such that if $n\gt N$ then $|\cos n\theta|\lt \frac{1}{10}$.
Now pick such an $n$, and consider $\cos(2n\theta)$. This is $2\cos^2(n\theta)-1$, which has absolute value $\gt \frac{98}{100}$. Since $2n\gt N$, we have reached a contradiction.
The question is harder than it might appear.
You ask "if the [something] is always moving in between $1$ and $-1$ shouldn't the summation equal $0$?" So consider: $$ 1 + (-1) + 1 + (-1)+\cdots. $$
The sum of the series is defined as the limit of the sequence of partial sums (or subtotals), and that sequence keeps alternating between $1$ and $0$, so it has no limit.
Sometimes one considers the limit of the average of the first $n$ partial sums, and that does indeed go to $0$.
But how does one know that $\cos(n\theta)$ actually keeps going back and forth between $1$ and $-1$? That's the hard part. It has a subsequence that approaches $1$; it has another subsequence that approaches $-1$; and for any number you pick between those two, it has a subsequence approaching that number. Proving that is not as easy as anything that can be done by methods used in courses where these concepts are first introduced. Unless you want a very long proof.
Hint: Show that for all $\epsilon > 0 $ and all integers $N$, there exists an integer $n> N $ such that $|\cos n \theta| > 1 - \epsilon$.
This shows that $a_n$ does not converge to 0, which is what you require.
Here the $\theta$ is fixed. So we can take $\theta =1$ for example. Now if the series is convergent, the $n$-th term must go to zero. But it does not.. So it is divergent..
