Applying Geometric Test for Infinite Series: $\sum_\limits{n=1}^\infty [\frac{1+\sin(n)}{10^n}]$ Given the following series:

$$\sum_\limits{n=1}^\infty [\frac{1+\sin(n)}{10^n}]$$

I am supposed to find whether it is divergent or convergent. My Original plan was to use limit comparison.
My Original Plan

$$\lim_\limits{n \to \infty}\frac{\frac{1+\sin(n)}{10^n}}{\frac{\sin(n)}{10^n}}$$

Which leads me to the following:

$$\lim_\limits{n \to \infty}(\frac{1}{\sin(n)}+1)$$

If the limit comparison fails does that mean its divergent, because the limit, in this case, does not exist?
Which thus lead me to test another comparison:
My Second Comparison
I used the second limit comparison, and got the following:

$$\lim_\limits{n \to \infty}(\frac{\frac{1+\sin(n)}{10^n}}{\frac{1}{10^n}})$$

Which leads me to the following, and it still does not exist:

$$\lim_\limits{n \to \infty}(1+\sin(n))$$

I am doing this from a section of a text book, and I can only use two methods, the Integral Test, and the Limit Comparison Test, and Geometric Test.
 A: Just for your curiosity.
Since you already received good answers about the convergence, let me show that we could get the exact value of the summation.
$$\sum\limits_{n=1}^{\infty}\frac{1+\sin(n)}{10^n}=\sum\limits_{n=1}^{\infty}\frac{1}{10^n}+\sum\limits_{n=1}^{\infty}\frac{\sin(n)}{10^n}=\frac19+\sum\limits_{n=1}^{\infty}\frac{\sin(n)}{10^n}$$
Now consider 
$$\sum\limits_{n=1}^{\infty}\frac{e^{in}}{10^n}=\sum\limits_{n=1}^{\infty}{e^{n(i-\log(10))}}=\sum\limits_{n=1}^{\infty}\Big[e^{i-\log(10)}\Big]^n$$ and we face another geometric series
$$\sum\limits_{n=1}^{\infty}\Big[e^{i-\log(10)}\Big]^n=\frac{e^i}{10-e^i}$$
$$\Im\left(\frac{e^i}{10-e^i}\right)=\frac{10 \sin (1)}{\sin ^2(1)+(10-\cos (1))^2}=\frac{10 \sin (1)}{101-20 \cos (1)}$$
$$\sum\limits_{n=1}^{\infty}\frac{1+\sin(n)}{10^n}=\frac 19+\frac{10 \sin (1)}{101-20 \cos (1)}$$
Making it more general, the same approach would give
$$\sum\limits_{n=1}^{\infty}\frac{\sin(an)}{b^n}=\frac{b \sin (a)}{b^2+1-2 b \cos (a)}\qquad \text{if} \qquad b >1$$
A: But you can use comparison test:
$$\sum_{n=1}^{\infty}\frac{1+\sin{n}}{10^n}<\sum_{n=1}^{\infty}\frac{2}{10^n}=2 \sum_{n=1}^{\infty}10^{-n}$$
Since it's bounded by power series, it's convergent itself.
A: The most straightforward method is to use direct comparison and then geometric series.
$\sum\limits_{n=1}^{\infty}a_n=\sum\limits_{n=1}^{\infty}\frac{1+\sin(n)}{10^n}\le \sum\limits_{n=1}^{\infty}\frac{2}{10^n}=\frac{2\left(\frac{1}{10}\right)}{1-\frac{1}{10}}=\frac{2}{9}$
