# How to determine if a series is convergent or divergent?

I have:

the series $\sum_{n=0}^{\infty} \frac{(2+\sin n)}{5^n}$.

I have split this up into $\sum_{n=0}^{\infty} \frac{2}{5^n} + \sum_{n=0}^{\infty} \frac{\sin n}{5^n}$. I know the first part is convergent by using geometric, but I am not sure how to approach the second part. Please help, thank you!

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Splitting turns out to be harmless, since one can prove that the second series converges. However, it is not a good idea. Definitely easier is comparison with $\sum_1^\infty \frac{3}{5^n}$. –  André Nicolas Nov 7 '12 at 6:23

There are many ways to show convergence/divergence of a series. Try the ratio test, the root test, the comparison test, or the integral test just to name a few. See here or here for a more thorough listing of tests for convergence.

In this particular case, the comparison test will be best. It may help to note that $-1 \leq \sin(n) \leq 1$ so that $1 \leq 2 + \sin(n) \leq 3$.

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@littleO. It will be easiest here by far. I was providing other tests as the OP seemed to be asking a general question about determining convergence. –  JavaMan Nov 7 '12 at 5:24
can you be a little more specific in explaining this? i have trouble figuring out which test i have to use –  Jaden Q Nov 7 '12 at 5:27
@JadenQ: On what specifically would like me to elaborate? –  JavaMan Nov 7 '12 at 5:28
i need help on figuring out the whole series, like how should i approach each part to know if they are either converge or diverge –  Jaden Q Nov 7 '12 at 5:32
@JadenQ: This is too general a question to answer (and perhaps there is no general answer). In short, it's just practice. We can see that the denominator grows very quickly, and the numerator is bounded between $1$ and $3$. Thus, we can compare the series to $\sum \frac{c}{5^n}$. –  JavaMan Nov 7 '12 at 5:37