The series is :$\sum_ {n = 0}^{\infty} \frac{\sin n }{ 5^n} $
I am not sure how to determine if this series is convergent or not. Can someone show me how?
Thanks
The series is :$\sum_ {n = 0}^{\infty} \frac{\sin n }{ 5^n} $
I am not sure how to determine if this series is convergent or not. Can someone show me how?
Thanks
The maximum range for $\sin(n)$ is $1$, so focus on $\frac{1}{5^n}$. If you take the limit of $\frac{1}{5^n}$ as $n$ goes to infinity, it converges to $0$ because the denominator gets larger and larger as it $n$ goes to infinity.
We have $|\sin n|\le 1$. Now compare with $\sum_0^\infty \frac{1}{5^n}$.
We conclude that $\sum_{n=0}^\infty \frac{|\sin n|}{5^n}$ converges. Thus the original series converges absolutely and therefore converges.
$\sum_{n=0}^{\infty} |\frac{\sin n}{5^n}| \le \sum_{n=0}^{\infty}\frac{1}{5^n}= \sum_{n=0}^{\infty}(\frac{1}{5})^n=\frac{1}{1-\frac{1}{5}}=\frac{5}{4}$
This series is majorized by $5^{-n}$, i.e. $\sum_{n=0}^\infty \left\vert \frac{\sin n}{5^n} \right\vert \leq \sum_{n=0}^\infty \frac{1}{5^n},$ and this latter series converges as a geometric series. So our original series converges by comparison.