For what positive integer values of $k$ make $\sin^2(k)=0$? Consider the infinite sum of $\frac{k\sin^2(k)}{k^2+1}$ whenever $k$ is a natural number with $k=1$ as the lower bound. I'm attempting to show that the series converges or diverges using the limit comparison test. However, my main concern is when $\sin^2(k) = 0$. For what positive integer values of $k$ make $\sin^2(k)=0$?
 A: $\sin(k) = 0$ when $k = A \pi$ where $A$ is an integer.
$A=k=0$ is the trivial solution. Let's show that it is unique.
If we look for an $A$ satisfying the solution, we must write
$k = A \pi$ and then  $\pi = k / A$ if $A \neq 0$
But we know that $\pi$ is not rational and that $k$ and $A$ may not be integers. Then the only solution is $0$.
A: As for convergence, hints: Your series converges iff $\sum (\sin^2 k)/k$ converges. Note that
$$\frac{\sin^2(k)}{k} + \frac{\sin^2(k+1)}{(k+1)} \ge \frac{\sin^2(k)+\sin^2(k+1)}{(k+1)}.$$
If you show that the function $\sin^2 (x) + \sin^2(x+1)$ has a positive minimum on $\mathbb R,$ you'll be done.
A: Your main concern shouldn't be when $\sin (k)$ is zero (and the ratio test doesn't work here). It should be that it doesn't approach zero 'too often'. In the present case it suffices to look at two consequtive values of say $k$ and $k+1$ and show that for at least one of them there is a lower bound on the value of $\sin^2$. You will then be able to show that the series diverge.
An interesting alternative problem (sorry, should probably not put it here) is if you replace $\sin^2(k)$ by $\sin(k^2)$, is the series conditionnally convergent (I guess yes, but have no idea of how to prove that).
A: All you need to know is what makes $sin(k)=0$. 
We know that on the unit circle that the value of $sin (k)$ is equal to the y-coordinate.  Hence, $sin(k)= 0$ at $k = 0$ and $k = nπ$—and at all angles coterminal with them.  In other words,
$sin (k) = 0$  when  $k = nπ$.
Radians can be converted to degree measures, which can be positive integers in this case. 
If you are looking for answers in radians, then $k=0$ is the only solution.
If you are looking for a positive answer, then there is no solution.
