# radius of convergence and the interval of convergence for $\sum_{n=0}^{\infty}(\frac{\pi}{2} +sin(n))x^n$

I'm having trouble finding the radius of convergence and the interval of convergence of the following sequence:

$$\sum_{n=0}^{\infty}(\frac{\pi}{2} +sin(n))x^n$$

I tried using the ratio test, but got stuck when trying to solve the limit.

Then went on to try the root test, thought that I could use the sandwich therom and therefore the limit is 1, which means that the radius is 1? Not sure (about the limit). If so, I need to check the ends of the interval: $$x=1, x=(-1)$$

for $$x=1$$ $$\sum_{n=0}^{\infty}(\frac{\pi}{2} +sin(n))1^n$$ = $$\sum_{n=0}^{\infty}(\frac{\pi}{2} +sin(n))$$

for $$x=(-1)$$ $$\sum_{n=0}^{\infty}(\frac{\pi}{2} +sin(n))(-1)^n$$

Both diverge (how do I prove that? and is that even true?), therefore the interval of convergence is $$(-1,1)$$?

Could really use help here.

Thanks!

It seems like you are confused about the boundary of your interval of convergence. Keep in mind that if $$\sum^{\infty}_{n=1}a_{n}$$ converges then $$a_{n} \to 0$$. This should remove all doubts. Since

{$$(-1)^{n}(\frac{\pi}{2}+sin(n))$$} and {$$(1)^{n}(\frac{\pi}{2}+sin(n))$$} are both bounded away from 0.