# 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 $\sin^2(k)=0$?

• You know the answer to that. Think about it: Where is $\sin x =0?$
– zhw.
Commented Aug 14, 2016 at 16:13
• Only $k=0{}{}{}{}$. Commented Aug 14, 2016 at 16:14
• Yes, I considered the case where $k=0$, but I wasn't sure if such $k$ was unique. But now I see that it is the only natural number satisfying the equation, because if another natural number $n$ existed such that $\sin^2(n)=0$, then $n=0$. Thanks! Commented Aug 14, 2016 at 16:18
• Assuming radians and not degrees. sin x = 0 if $x = k\pi$. $\pi$ is irrational. So..... Commented Aug 14, 2016 at 16:31

$\sin(k) = 0$ 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 $0$.

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.

• and the last thing to do is very easy. Nice approach !
– user354674
Commented Aug 14, 2016 at 16:34

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).

• Indeed I am not answering the question. But looking ahead, the question is not relevent to the problem (convergence or divergence of the series), so I am just indicating how to solve the underlying problem. And I see that zhw has taken up this thread... Commented Aug 14, 2016 at 16:36

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.

• Not quite, we also need to know that $\pi$ is irrational. Commented Aug 14, 2016 at 16:20
• We can't automatically assume that in this case?
– 关一骏
Commented Aug 14, 2016 at 16:22
• Update: Benedict is asking for a positive value, but 0 isn't a positive integer; I'm not very certain of how to solve this problem any more.
– 关一骏
Commented Aug 14, 2016 at 16:23
• Yes, but the the irrationality of $\pi$ is the crucial part. Commented Aug 14, 2016 at 16:23
• Why a problem ? This simply means that no such $k$ exists. The question states clearly enough that the OP wants a positive integer. Commented Aug 14, 2016 at 16:26