When can a sum be well approximated by an integral? The problem I'm struggling with goes as follows:
Show, that the sum $\sum_{n=0}^∞ e^{-\frac{n(n+1)k}{c}}(2n+1)$ can for $c\gg k$ be well approximated by an integral and compute that integral.
My intuition tells me that if the terms in the exponential are close enough together, the sum behaves just like Riemann integral, however since the $n$s are growing as a square, there will for a sufficiently large $n$ be a term that outweighs the condition on the constants and the terms get artificially far away from each other. For the largest $n$s, the exponential approaches 0 quickly, but what happens "in the middle"?
$\frac{\text{d}}{\text{d}x} e^{-\frac{x(x+1)k}{c}}(2x+1) \rightarrow 2$ for large $c$, so the argument for a flat function doesn't seem viable...
 A: Too long for a comment.
The heuristic considerations can be useful for finding a solution to the problem.
In your case $S(\alpha)=\sum_{n=0}^∞ e^{-\alpha n^2-\alpha n}(2n+1)$, where $\alpha=\frac{k}{c}<<1$ - a small parameter, so we can get asymptotics based on this small $\alpha$.
We see that at $n\sim\frac{1}{\sqrt\alpha}>>1$ $\,\,\alpha n^2\sim 1$, but $\alpha n\sim\sqrt{\alpha}<<1$. If $n\sim\frac{1}{\alpha}$, $\,\alpha n\sim 1$ and $\alpha n^2>>1$.
So, the second term in the exponent power contributes only when the exponent is already $<<1$ due to the first term. Therefore we can omit the second term while finding the leading asymptotics.
$$S(\alpha)=\frac{1}{\alpha}\sum_{n=0}^∞ e^{-\alpha n^2-\alpha n}(2n\alpha+\alpha)\approx\frac{1}{\alpha}\sum_{n=0}^∞ e^{-\alpha n^2}(2n\alpha)\approx\frac{1}{\alpha}\sum_{n=0}^∞ e^{-\alpha n^2}\alpha\Delta(n^2)$$
where $\Delta(n^2)\approx (n+1)^2-n^2$.
$$S(\alpha)\approx\frac{1}{\alpha}\sum_{n^2=0}^∞ e^{-\alpha n^2}\alpha\Delta(n^2)\approx\frac{1}{\alpha}\int_0^\infty e^{-x}dx=\frac{1}{\alpha}$$
To find asymptotics power series of $\alpha$ we can use the Euler-Maclaurin formula:
$$\sum_{n=a}^b f(n) =\int_a^bf(x)dx+\frac{1}{2}(f(a)+f(b))+\frac{1}{12}(f^{(1)}(b)-f^{(1)}(a))+\frac{B_4}{4!}(f^{(3)}(b)-f^{(3)}(a))+...$$
It is important that every new term in the series declines as additional powers of $\alpha$.
In our case $f(x)=e^{-\alpha (x^2+x)}(2x+1)$, $a=0$ and $b=\infty$, so all calculations are straightforward. Integral gives the main asymptotics; second and third terms contribute as $O(1)$ with their main terms.
Let's evaluate, for example, $\frac{B_4}{4!}(f^{(3)}(b)-f^{(3)}(a))$, where $B_4=-\frac{1}{30}$ - Bernoulli number. Taking derivatives and limits ($x=0$ and $x=\infty$) we get
$$\frac{1}{4!30}f^{(3)}(0)=-\frac{12\,\alpha}{4!30}+O(\alpha^2)$$
