Proof of sum in an inequality I was having hard time solving this one, any help will be greatly appreciated.
prove that:
$$
{39\over e^2}\le\sum_{n=1}^\infty {4n^2-1\over e^n}-{3\over e}\le{54\over e^2}
$$
 A: Hint:
Let $f(x)=\dfrac{4x^2-1}{e^x}$. Note that this function decreases for $x \ge 3$.
Further, we can check that $f(2)> \int_2^3 f(x) dx$. 
$$\implies \int_{2}^3 f(x)dx +\int_{3}^\infty f(x)dx < f(2)+\sum_{3}^\infty f(n) < f(2)+\int_{2}^\infty f(x)dx $$
$$\implies \int_{2}^\infty f(x)dx < \sum_{2}^\infty f(n) < f(2)+\int_{2}^\infty f(x)dx $$
$$ \implies \frac{39}{e^2} < \sum_{2}^\infty f(n) < \frac{54}{e^2} $$
A: $4\displaystyle \sum_{n=1}^\infty n^2\left(\frac{1}{e}\right)^n$ is related to the infinite series: $\displaystyle \sum_{n=1}^\infty x^n, |x| < 1$ through twice differentiation. Can you continue?
A: We know $\sum_{n=0}^\infty x^n=\frac{1}{1-x}$.  Taking derivatives $\sum_{n=1}^\infty nx^{n-1}=\frac1{(1-x)^2}$.  Multiply both sides by $x$ and take another derivative to get $\sum_{n=1}^\infty n^2x^{n-1}=\frac{1-x^2}{(1-x)^4}$.  Thus $\sum_{n=1}^\infty 4n^2x^n=\frac{4x(1-x^2)}{(1-x)^4}$.  Thus $\sum_{n=1}^\infty(4n^2x^n-x^n)=\frac{4x(1-x^2)}{(1-x)^4}-\sum_{n=1}^\infty x^n=\frac{4x(1-x^2)}{(1-x)^4}- \frac{x}{1-x}$.  Thus $\sum_{n=1}^\infty(4n^2-1)x^n=\frac{4x(1-x^2)}{(1-x)^4}- \frac{x}{1-x}$.  Plug in $\frac1e$ to get $\sum_{n=1}^\infty\frac{4n^2-1}{e^n}=\frac{4e(e+1)}{(e-1)^3}-\frac1{e-1}$.
