<s>[INST] Proof of sum in an inequality I was having hard time solving this one, any help will be greatly appreciated.
prove that:
$$
{39\over e^2}\le\sum_{n=1}^\infty {4n^2-1\over e^n}-{3\over e}\le{54\over e^2}
$$
 [/INST] 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 $\sum_{n=1}^\infty\frac{4n^2-1}{e^n}=\frac{4e(e+1)}{(e-1)^3}-\frac1{e-1}$.
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