Using Limit Comparison Test 
Say I have this function:
$$\sum_{n=1}^{\infty} \frac{9^n}{3 + 10^n}$$
I think it's similar to:
$$\sum_{n=1}^{\infty} \frac{9^n}{10^n}$$
which is similar to a geometric series with $r = \frac{9}{10}$ which converges.
How do I show this? I'm stuck here:
$$\lim_{n \to \infty} \frac{\frac{9^n}{3 + 10^n}}{\frac{9^n}{10^n}}$$
 A: We begin with the fraction you state in the problem, and simplify, cancelling out the $9^n$ factor:
$$\frac{9^n / (3+10^n)}{9^n / 10^n} = \frac{10^n}{10^n+3}$$
As $n\to\infty$, it should be clear that the $+3$ becomes more and more "negligible," that the $10^n$ dominates the growth of the fraction. Thus, in the infinite limit, we can effectively ignore the $+3$ and claim:
$$\lim_{n\to\infty} \frac{9^n / (3+10^n)}{9^n / 10^n} = \lim_{n\to\infty} \frac{10^n}{10^n+3} = \lim_{n\to\infty} \frac{10^n}{10^n} = 1$$
A: In your example, $a_{n}=\frac{9^{n}}{3+10^{n}}, b_{n}=\frac{9^n}{10^n}, n\geq 1$
Note that, $\frac{a_{n}}{b_{n}}=\frac{10^{n}}{3+10^{n}}$ and $a_{n}>0,b_{n}>0$ for all $n\in \mathbb{N}$
So, $\lim\limits_{n \to \infty}\frac{a_{n}}{b_{n}}=\lim\limits_{n \to \infty}\frac{10^{n}}{3+10^{n}}=\lim\limits_{n \to \infty}\frac{1}{\frac{3}{10^{n}}+1}=1$ (because $\lim\limits_{n \to \infty} \frac{3}{10^{n}}=0$)
Therefore, in the above theorem we get $c=1>0$
Since the series $\sum_{n=1}^{\infty}b_{n}$ converges(since it is a geometric series with common ratio $\frac{9}{10}<1)$, by the theorem you provided, $\sum_{n=1}^{\infty}a_{n}$ also converges.
