Prove that $\sum_{k=1}^{\infty}\frac{k^2}{e^k}< \sum_{k=1}^{\infty}\frac{k}{2^k}$

Question

Prove that :

$$\sum_{k=1}^{\infty}\frac{k^2}{e^k}< \sum_{k=1}^{\infty}\frac{k}{2^k}$$

Without calculating the value of these series .

The partial sum formula are :

$$\sum_{k=1}^n \frac{k}{2^k} = 2^{-n}(-n+2^{n+1}-2)$$

$$\sum_{k=1}^n \frac{k^2}{e^k} =\frac{ (e^{-n} (-e^2 n^2 + 2 e n^2 - n^2 - 2 e^2 n + 2 e n + e^{n + 1} + e^{n + 2} - e^2 - e))}{(e - 1)^3}$$

I have tried a straightforward comparison like :

$$\frac{x^2}{e^x}\leq \frac{x}{2^x}$$

Wich is true for $x\geq 6$

So we have a problem with the first terms .

If you have a trick (maybe using integral) or a good answer I take .

Any helps is appreciated.

Thanks a lot!

Answer 1

Observe

$$ \sum_{k=1}^{11}(\frac{k^2}{e^k}-\frac{k}{2^k}) = -0.002905 $$

Further, include your finding that for $x > 6$, $$ \frac{x^2}{e^x}- \frac{x}{2^x} \leq 0 $$

This shows that the partial sum for more than 11 terms will always obey the required inequality, hence also the infinite sum.

Answer 2

Thsi is not very rigorous.

Let $$a_n=\frac{\sum_{k=1}^{n}\frac{k^2}{e^k}} { \sum_{k=1}^{n}\frac{k}{2^k}}$$ Using the partial sums you gave $$a_n=\left(\frac{2}{e}\right)^n \frac{-e^2 n^2+2 e n^2-n^2-2 e^2 n+2 e n+e^{n+1}+e^{n+2}-e^2-e}{(e-1)^3 \left(-n+2^{n+1}-2\right)}$$ The numerator is basically driven by $\left(e^{n+1}+e^{n+2}\right)$ and the denominator is almost $(e-1)^3\, 2^{n+1}$. So, at he limit $$a_n \sim \left(\frac{2}{e}\right)^n \frac{e^{n+1}+e^{n+2}}{(e-1)^3\, 2^{n+1}}=\frac{e (e+1)}{2 (e-1)^3}\approx 0.996147$$