How to prove that $\sum\limits_{k=0}^{\infty}(k+1) e^{-\frac{k^2}{2}} < 3$?

Question

What could be a possible approach to find the sum of this series $\sum\limits_{k=0}^{\infty}(k+1) e^{-\frac{k^2}{2}}$ (the sum or an estimate of the sum)? Any answer will be appreciated. Thanks.

Answer 1

It's a little messy but not too hard to show that

$${4\over e^{9/2}}+{5\over e^{16/2}}+{6\over e^{25/2}}+\cdots={4\over e^4}\left({1\over e^{1/2}}+{5/4\over e^4}+{6/4\over e^{17/2}}+\cdots \right)\lt{4\over e^4}\lt{4\over40}={1\over10}$$

It follows that

$$1+{2\over e^{1/2}}+{3\over e^2}+{4\over e^{9/2}}+\cdots\lt1+{2\over\sqrt2}+{3\over7}+{1\over10}={107\over70}+\sqrt2\lt3$$

where the final inequality can be verified from the equivalence

$${107\over70}+\sqrt2\lt3\iff107+70\sqrt2\lt210\iff70\sqrt2\lt103$$

and the easy calculation

$$(70\sqrt2)^2=9800\lt10{,}000=100^2\lt103^2$$

Remarks: We use the inequalities $2\lt e$ and $7\lt e^2$ (so that $40\lt e^4$) in obvious spots in the comparisons. These also help in the "messy" comparison

$${1\over e^{1/2}}+{5/4\over e^4}+{6/4\over e^{17/2}}+\cdots\lt{1\over e^{1/2}}\left(1+{1\over e^2}+{1\over e^4}+\cdots \right)={1\over\sqrt e}{1\over1-{1\over e^2}}\lt{1\over\sqrt2}{1\over1-{1\over7}}={7\over6\sqrt2}\lt1$$

A more challenging problem might be to show that

$$\sum_{k=0}^\infty(k+1)e^{-k^2/2}\lt e$$