Comparison / p-series problem 
*

*Given $\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)}$ does it converge or diverge?

*Given $\sum_{n=1}^{\infty}\frac{5^{n}+4^{n}}{7^{n}-2^{n}}$ does it converge or diverge?


For 1. I know that $\sum_{n=1}^{\infty}\frac{1}{n(n+1)} \rightarrow 1$. Therefore the first series converges by comparison test (since in 1 the terms are smaller.)
For 2. It seems a general strategy for exponential series is to try and compare it to a geometric series. Dividing by the highest exponential I get $\sum_{n=1}^{\infty}\frac{(5/7)^{n} + (4/7)^{n}}{1 - (2/7)^{n}}$. I get stuck now since the denominator is being made smaller, I think this might diverge, but I'm not sure what to compare it too. 
Thanks all. A few different methods for solving this. I'll also take a look at asymptotic comparison test.
 A: Note that as $n\to\infty$ we have
\begin{align}
&n(n+1)(n+2)\sim n^3&
&5^n+4^n\sim 5^n&
&7^n-2^n\sim 7^n
\end{align}
hence
\begin{align}
&\frac 1{n(n+1)(n+2)}\sim\frac 1{n^3}&
&\frac{5^n+4^n}{7^n-2^n}\sim\left(\frac 57\right)^n
\end{align}
hence asymptotic comparison test applies.
A: For the second:


*

*$5^{n}+4^{n} < 2 \cdot 5^n$

*$7^{n}-2^{n} > \frac12 \cdot 7^n$

*$\dfrac{5^{n}+4^{n}}{7^{n}-2^{n}} < 4\cdot \left(\dfrac57\right)^n$

*$\sum\limits_{n=1}^{\infty}\dfrac{5^{n}+4^{n}}{7^{n}-2^{n}} < 4\cdot \sum\limits_{n=1}^{\infty}\left(\dfrac57\right)^n = 4 \cdot \dfrac52 =10$
so it converges, in fact to just under $4.44$
A: For the second series, notice that:
$$\frac{5^{n}+4^{n}}{7^{n}-2^{n}} = \frac{5^{n}\left(1+\left(\frac{4}{5}\right)^{n}\right)}{7^{n}\left(1-\left(\frac{2}{7}\right)^{n}\right)}.$$
It is clear that:


*

*$1+\left(\frac{4}{5}\right)^{n} \leq 1 + \frac{4}{5} = \frac{9}{5} ~\forall n \in \mathbb{N}.$ 

*$1-\left(\frac{2}{7}\right)^{n} \geq 1 - \frac{2}{7} = \frac{5}{7} ~\forall n \in \mathbb{N}.$
As a consequence:
$$\frac{5^{n}\left(1+\left(\frac{4}{5}\right)^{n}\right)}{7^{n}\left(1-\left(\frac{2}{7}\right)^{n}\right)} \leq \frac{5^{n} \cdot \frac{9}{5}}{7^{n} \cdot \frac{5}{7}} = \frac{63}{25} \cdot \left(\frac{5}{7}\right)^n \sim \left(\frac{5}{7}\right)^n.$$
