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
 A: You could also consider $$S_k=\sum_{n=0}^k n(n+1)x^n$$ and rewrite $$n(n+1)=n(n-1)+2n$$ making $$S_k=\sum_{n=0}^k n(n-1)x^n+2\sum_{n=0}^k nx^n=x^2\sum_{n=1}^k n(n-1)x^{n-2}+2x\sum_{n=0}^k nx^{n-1}$$ that is to say $$S_k=x^2 \left(\sum_{n=0}^k x^{n} \right)''+2x\left(\sum_{n=0}^k x^{n} \right)'$$ and use $$\sum_{n=0}^k x^{n}=\frac{1-x^{k+1}}{1-x}$$ Compute the derivatives and, at the end, make $x=\frac 14$. 
A: One way to approach it is to realize that, since you have a polynomial divided by an exponential, the sequence goes very quickly to $0$. Of course, approaching zero is not a sufficient condition, but the key word here is "quickly".
More precisely, it's easy to prove that for large enough values of $n$ it holds that $n(n+1)\leq 4^{n/2}$, therefore we can write the sequence as
$$
\frac{n(n+1)}{4^n} = \left(\frac{n(n+1)}{4^{n/2}}\right)\cdot \frac{1}{4^{n/2}}
$$
which is less than $4^{-n/2}$ because of what I argued initially. Given this, our sequence tends to zero at least as fast as an exponential and therefore, the series converges (this last arguments relies on the geometric series).
A: Why don't you use the ratio test? 
$$
\frac{|a_{n+1}|}{|a_n|}=\frac{\frac{(n+1)(n+2)}{4^{n+1}}}{\frac{n(n+1)}{4^n}}=\frac14\cdot\frac{n+2}{n}=\frac14\cdot\frac1{1+\frac2n}\to \frac14<1.
$$
The ratio test yields that $\sum a_n$ congerves.
A: You can use a limit comparison test to say $\sum_{n=1}^{\infty} \frac{1}{n^2}$. We have
$$ \frac{\frac{n(n+1)}{4^n}}{\frac{1}{n^2}} = \frac{n^3(n+1)}{4^n} \to 0 $$
and since $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges, we get that $\sum_{n=1}^{\infty} \frac{n(n+1)}{4^n}$ also converges.
A: Hint:
Use asymptotic analysis: $n(n+1)=o(2^n)$, hence
$$\frac{n(n+1)}{4^n}=\frac1{2^{2n}}o(2^n))=o\Bigl(\frac1{2^n}\Bigr).$$
A: Hint:
$$\forall n \ge 5, \frac{n(n+1)}{4^n}<2^{-n}$$
A: 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 $x=\frac14$:
$$\sum_{n=1}^{\infty} \frac{n(n+1)}{4^n}=\frac{2\cdot \frac14}{(1-\frac14)^3}=\frac{32}{27}.$$
