Convergence or divergence of $ \sum \frac{n(n+1)}{4^n} $ 
Convergence or divergence of
  $$ \sum_{n=1}^{\infty} \frac{n(n+1)}{4^n} $$

Considering 
$$ a_n= \frac{n(n+1)}{4^n} \leq \frac{n(n+2)}{4^n} =b_n$$
As the integral test is conditioned on having a function positive and decreasing. $a_n$ is first increasing then decreasing. 
This $b_n$ is obviously not appropriate. How would you find an upper bound to as to do a comparison test?
What would be another approach ?
Much appreciated

The series is convergent (by the Root test: $\lim_\limits{n\to\infty} \frac{\sqrt[n]{n(n+1)}}{4}=\frac14<1)$ and its sum can be calculated. Consider the series (for $|x|<1$):
$$\sum_{n=0}^{\infty} x^{n+1}=x+x^2+x^3+x^4+\cdots=\frac{x}{1-x}.$$
Take its derivative:
$$\sum_{n=0}^{\infty} (n+1)x^n=1+2x+3x^2+4x^3+\cdots=\frac{1}{(1-x)^2}.$$
Take derivative again:
$$\sum_{n=1}^{\infty} n(n+1)x^{n-1}=2+6x+12x^2+\cdots=\frac{2}{(1-x)^3}.$$
Multiply both sides by $x$:
$$\sum_{n=1}^{\infty} n(n+1)x^n=2x+6x^2+12x^3+\cdots=\frac{2x}{(1-x)^3}.$$
Substitute $x=\frac14$:
$$\sum_{n=1}^{\infty} \frac{n(n+1)}{4^n}=\frac{2\cdot \frac14}{(1-\frac14)^3}=\frac{32}{27}.$$
