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.



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.


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