Evaluate $ \lim_{h \to 0} h\sum_{n=0}^{\infty} e^{-n^2h^2}$ Evaluate  
$$ \lim_{h \to 0} h\sum_{n=0}^{\infty} e^{-n^2h^2}$$ 
I think it is somehow related to Riemann Sums, but I'm not sure.
Please help.
 A: I think a good way is to use integral test for convergence.  
This shows that for all $h > 0$, you have :
$$|\sum_{n=0}^\infty e^{-n^2 h^2} - \int_0^\infty e^{-x^2 h^2} dx| \leq 1$$.
Using this you can see that 
$$\lim_{h \to 0} h \sum_{n=0}^\infty  e^{-n^2 h^2} = \lim_{h \to 0} h  \int_0^\infty e^{-x^2 h^2} dx$$.
This last integral is a Gaussian integral, and can be evaluated for any $h > 0$:
$$ \int_0^\infty e^{-x^2 h^2} dx = \frac{\sqrt{\pi}}{2} \frac{1}{h}$$ 
So finally $$\lim_{h \to 0} h \sum_{n=0}^\infty  e^{-n^2 h^2} = \frac{\sqrt{\pi}}{2}$$ 
A: If a function $f$ is Riemann-integrable on $[0,\infty)$, then its integral $\int_0^\infty f(x)\,dx$ is equal to
$$\lim_{h\rightarrow 0}\,h\sum_{n=0}^\infty f(nh)$$
Now you just have to set $f$ appropriately. See this link if you can't do the resulting integration.
Edited to add: As commenters have pointed out, I have over-simplified here. What I should have said was this:
Let $f:[0,\infty)\rightarrow \Bbb R$ be a function of bounded variation, and suppose that it is Riemann-integrable over $[0,M]$ for all $M\ge 0$, and that $\lim_{M\rightarrow\infty}\int_0^M f(x)\;dx$ exists. Then this limit, which we denote by $\int_0^\infty f(x)\,dx$, is equal to
$$\lim_{h\rightarrow 0}\,h\sum_{n=0}^\infty f(nh)$$
