Is $\sum_{n=1}^\infty \sin\left(\frac{n\pi}{4}\right)$ convergent? Is $$\sum_{n=1}^\infty \sin\big(\frac{n\pi}{4}\big)$$ convergent?
I'm pretty sure it isn't due to the oscillatory nature of the sine function but I'm not sure how I would prove it. Would using the definition of convergence of a series work? (I.e saying that the sequence of partial sums cannot converge since you can consider two sub sequences that converge to different limits? Example $n=0$, 4, 8, 12,... converges to 0 and $n=2$, 6, 10, 14... converges to $\frac{1}{2}$). Any ideas?
 A: Hint:
If $\sum_n a_n$ converges, then $\lim_n a_n = 0$.
To prove this result, let $s_n=\sum_{k=1}^n a_k$ be the $n^{th}$ partial sum. Then $s_n \to s$ for some $s$. Thus, given $\epsilon>0$ we can find $N \in \mathbb{N}$ such that $|s_n-s| < \frac{\epsilon}{2}$ if $n > N$. Thus if $n>N$, then $|a_{n+1}|=|s_{n+1}-s_n| \leq |s_{n+1}-s|+|s-s_n| < \frac{\epsilon}{2}+\frac{\epsilon}{2} = \epsilon$. So $a_n \to 0$.
A: HINT: (An alternative attempt.) Partial sums are periodic, nonconstant,indeed. Just count them to $n=16$ to see the behaviour.
A: A variant: consider the associated complex series $\displaystyle\sum\limits_{n=1}^\infty \mathrm e^{\tfrac{n\mathrm i\pi}4}$ (the O.P.'s series is its imaginary part).
It's a geometric series, and a partial sum is:
$$\sum\limits_{n=1}^N \mathrm e^{\tfrac{n\mathrm i\pi}4}=\smash{\frac{1-\mathrm e^{\tfrac{(N+1)\mathrm i\pi}4}}{1-\mathrm e^{\tfrac{\mathrm i\pi}4}}}$$
These partial sums can't have a limit since $\mathrm e^{\tfrac{(N+1)\mathrm i\pi}4}$ is ($4$-)periodic.
