Prove these bounds on $\sum_{n=1}^\infty \frac{4n^2 -1}{e^n}$

Question

I already found this series converge but I don't know how to prove this inequality: $$ \frac{39}{e^2} \le \left[\sum_{n=1}^\infty \frac{ (4n^2 -1)}{e^n}\right] - \frac{3}{e} \le \frac{54}{e^2} $$

All I've found out is that $\int_{2}^{\infty} \frac{ (4n^2 -1)}{e^n} = \frac{39}{e^2}$

And I don't know what to do next to prove it.

Thanks for your help!

your sum does converge and hase the sum $$\frac{-1+6 e+3 e^2}{(e-1)^3}$$ — Dr. Sonnhard Graubner, May 28 '17 at 13:56

score 2 · Answer 1 · answered May 28 '17 at 13:57

Hint. One may recall that (see here) $$ \sum_{n=1}^{\infty}x^n=\frac{x}{1-x}, \qquad \sum_{n=1}^{\infty}n^2x^n=\frac{x^2+x}{(1-x)^3}, \qquad |x|<1, $$ giving, by putting $x=\dfrac 1e$, $$ \sum_{n=1}^\infty \frac{ (4n^2 -1)}{e^n}=\frac{3e^2+6e-1}{(e-1)^3}=\color{blue}{\frac3e}+\color{red}{\frac{15e^2-10e+3}{e(e-1)^3}}. $$

Prove these bounds on $\sum_{n=1}^\infty \frac{4n^2 -1}{e^n}$

1 Answers1