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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

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1  
This seems the same question as your previous post: math.stackexchange.com/questions/231935/… –  JavaMan Nov 7 '12 at 6:32

4 Answers 4

up vote 4 down vote accepted

$\sum_{n=0}^{\infty} |\frac{\sin n}{5^n}| \le \sum_{n=0}^{\infty}\frac{1}{5^n}= \sum_{n=0}^{\infty}(\frac{1}{5})^n=\frac{1}{1-\frac{1}{5}}=\frac{5}{4}$

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you have explained in a very detailed way, thanks so much –  Jaden Q Nov 7 '12 at 6:33
    
You're welcome. –  The Substitute Nov 7 '12 at 6:34
2  
Using the formula for the geometric series, we can actually, compute the sum: $$ \begin{align} \sum_{n=0}^\infty\frac{\sin(n)}{5^n} &=\mathrm{Im}\left(\sum_{n=0}^\infty\frac{e^{in}}{5^n}\right)\\ &=\mathrm{Im}\left(\frac1{1-e^i/5}\right)\\ &=\mathrm{Im}\left(\frac5{5-e^i}\frac{5-e^{-i}}{5-e^{-i}}\right)\\ &=\mathrm{Im}\left(\frac{25-5e^{-i}}{26-10\cos(1)}\right)\\ &=\frac{5\sin(1)}{26-10\cos(1)}\\ &\stackrel.=0.20427050707618 \end{align} $$ –  robjohn Nov 7 '12 at 12:08

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.

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Not that it really matters, but the sum actually starts at $n = 0$ –  sidht Nov 7 '12 at 6:21
    
thanks! it helps –  Jaden Q Nov 7 '12 at 6:25
    
@jak: It certainly doesn't matter, but one might as well use the same beginning as the OP. Thanks! –  André Nicolas Nov 7 '12 at 6:26
    
Glad someone mentioned that AC $\implies$ C theorem. It's a handy tool. –  Joe Nov 7 '12 at 7:05

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.

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thanks it helps –  Jaden Q Nov 7 '12 at 6:29

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.

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