# Radius of Convergence of $\sum ( \sin n) x^n$.

Thank you very much in advance for any assistance/advice on solving this problem. I am fairly new to power series and determining the radius of convergence.

Determine, with proof, the radius of convergence of $\sum ( \sin n) x^n$.

My understanding of this problem is that we cannot employ the ratio test since it will not really give us something we can use to determine the convergence (actually, I don’t think it even applies). One approach that I can think of is to compare this series with another one, but could it involve the integral test since $\sin n$ is continuous? Also, my book does not include the lower and upper limits of the sum, but I am assuming they go from 0 to $\infty$.

I am using the textbook Introduction to Analysis by Arthur Mattuck.

#### Solutions Collecting From Web of "Radius of Convergence of $\sum ( \sin n) x^n$."

Since $|\sin n|\le 1$ for all $n$, our series converges absolutely for any $x$ with $|x|\lt 1$, by comparison with the geometric series $\sum x^n$.

Thus the radius of convergence is $\ge 1$. It remains to show that the radius is not $\gt 1$.

To do this, it is enough to show that for any $\delta\gt 0$, the series
$\sum \sin n (1+\delta)^n$ does not converge.

We do this by showing that the terms $\sin n (1+\delta)^n$ cannot have limit $0$.

The numbers $\sin n$ are dense in the interval $[-1,1]$. In particular, there are infinitely many $n$ such that $|\sin n|\gt 1/2$. This completes the proof.

Remark: We can prove the result without referring to the fact that the numbers $\sin n$ are dense in $[-1,1]$. We show that $\sin n$ cannot have limit $0$. For suppose to the contrary that $|\sin n|$ is close to $0$ for all large enough $n$. Suppose for instance that for all $n\gt N$, we have $|\sin n|\lt 1/10$. Consider $\sin(n+1)=\sin n \cos 1+\cos n\sin 1$. The term $\sin n \cos 1$ has absolute value $\lt 1/10$. The term $\cos n\sin 1$ has absolute value $\ge \sqrt{1-1/100}\sin 1\gt 0.8$, and therefore $|\sin(n+1)|\gt 0.7$.

Hint: You can use the root test.