Is the following series converging or diverging. $\sum_{n=1}^{\infty}\dfrac{n+4^n}{n+6^n}$ I know one solution. 
That is by Doing comparison with $\dfrac{4^n+4^n}{6^n}$
Wondering if there are more ways to do it
 A: Comparison is good. Most students would automatically go to the Ratio Test. That also works well, but to me is less attractive. 
A: By the ratio test, we have
$$
r=\lim_{n \to +\infty}\biggl|\frac{a_{n+1}}{a_n}\biggr| = \lim_{n \to +\infty}\frac{n+1 + 4^{n+1}}{n + 4^n}\cdot \frac{n + 6^{n}}{n + 1 + 6^{n+1}} 
$$
$$
= \lim_{n \to +\infty}\frac{\frac{n+1}{4^n}+ 4}{\frac{n}{4^n} + 1}\cdot \frac{\frac{n}{6^n}+ 1}{\frac{n+1}{6^n} + 6} = \frac{2}{3} < 1
$$
The series converge.
A: It is good to get good knowledge of the size of the terms involved.
For instance, you may have noticed that $4^n$ is way bigger than $n$.
You can elaborate as follows:
$$n+4^n=4^n(\frac{n}{4^n}+1),$$ and note that the second factor tends to $1$. (This is what you do with rational functions of $n$,when you calculate limits).
If you proceed likewise with the denominator, you have the same thing. Thus, if you call your initial series $a_n$, you have that 
$a_n\sim (4/6)^n$, where this means that $lim\ \frac{a_n}{(2/3)^n}=1$.
Even if parameters would change, this solution will essentially work out.
