I have reduced the expression that I am working on to the following sum of series, which is definitely converging. It would be great if someone can help me out with this or suggest ways this can be done.
$\sum_\limits{i=0}^{\infty} (i+1)^{2}.e^{-(2i+1)^2}$
It might be possible to bound the above using the integral. Even a tight bound to the above would work for me.
EDIT : I am more interested in the following : $\sum_\limits{i=0}^{\infty} (i+1)^{2}.e^{-\alpha(2i+1)^2}$ where $\alpha$ can be any constant (known). I am making this edit because for a constant multiplied the estimation approach by simply calculating a first few values won't work. In this case one would have to calculate first few values for different values of $\alpha$. Also, for different values of $\alpha$ it might not be able to always truncate after a fixed number. So, in essence, an analytical approach would be much more helpful.
$$e^{-1}+4e^{-9}+9e^{-25}+16e^{-49}\approx 0.36837308051278053458657911933771842$$ should be good enough.
The truncation error is
$$\sum_{i=4}^\infty(i+1)^2e^{-(2i+1)^2}=\sum_{i=4}^\infty(i+1)^2e^{-4i^2-4i-1}<\sum_{i=4}^\infty(i+1)^2e^{-64-4i-1}<2\cdot10^{-34}$$ as can be computed analytically.
You can obtain this estimate from $$\sum_{i=n}^\infty b^{i}=\frac{b^{n}}{1-b},$$ $$\sum_{i=n}^\infty (i+1)b^{i}=\left(b\frac{b^{n}}{1-b}\right)',$$ $$\sum_{i=n}^\infty (i+1)^2b^i=\left(b\left(b\frac{b^{n}}{1-b}\right)'\right)'.$$