A step in zeta function regularization I'm just wondering about the mathematical step
$$\sum_{n=1}^\infty n\exp[-\epsilon n\sqrt x]=\frac1{\epsilon^2 x}-\frac1{12}+\mathcal O(\epsilon).$$
Why is this equality so? I see that
$$\sum_{n=1}^\infty n\exp[-\epsilon n\sqrt x]=-\frac1{\sqrt x}\frac{\partial}{\partial\epsilon}\frac{1}{1-\exp[-\epsilon\sqrt x]}\simeq-\frac1{\sqrt x}\frac{\partial}{\partial\epsilon}\frac{1}{\epsilon\sqrt x}=\frac1{\epsilon^2x}.$$
But how about the $-\frac1{12}$?
 A: \begin{align*}
\sum_{n=1}^\infty n\exp[-\epsilon n\sqrt x]
&=-\frac1{\sqrt x}\frac{\partial}{\partial \epsilon}\frac{1}{1-\exp[-\epsilon\sqrt x]}\\
&=\frac{\exp[-\epsilon\sqrt x]}{(1-\exp[-\epsilon\sqrt x])^2}\\
&=\frac{1}{\exp[\epsilon\sqrt x]-2+\exp[-\epsilon\sqrt x]}\\
&\simeq\frac{1}{\frac2{2!}(\epsilon\sqrt x)^2+\frac{2}{4!}(\epsilon\sqrt x)^4}\\
&=\frac{1}{\epsilon^2 x}\frac1{1+\frac1{12}\epsilon^2 x}\\
&\simeq\frac{1}{\epsilon^2 x}(1-\frac1{12}\epsilon^2 x)\\
&=\frac1{\epsilon^2 x}-\frac1{12}.
\end{align*}
A: The geometric series formula tells us
$$
\sum_{n=1}^{\infty} e^{-\epsilon n\sqrt{x}} = \frac{e^{-\epsilon \sqrt{x}}}{1-e^{-\epsilon\sqrt{x}}}.
$$
The derivative with respect to $\epsilon$ of the left hand side gives $-\sqrt{x}$ times your sum. Therefore your sum is equal to
$$
-\frac{1}{\sqrt{x}}\frac{\partial}{\partial\epsilon}\frac{e^{-\epsilon \sqrt{x}}}{1-e^{-\epsilon\sqrt{x}}}
$$
which I believe is equal to
$$
\frac{1}{4\sinh^2(\epsilon\sqrt{x}/2)}.
$$
At small $\epsilon$ this diverges like $\frac{1}{x\epsilon^2}$. 
Multiply it by $\epsilon^2$ to get something that is finite as $\epsilon\to 0$, do a Taylor series in $\epsilon$, and then divide back by $\epsilon^2$.
Doing that: Let
$$
f(\epsilon) = \frac{\epsilon^2}{4\sinh^2(\epsilon\sqrt{x}/2)}
$$
$f(\epsilon)$ is $\epsilon^2$ times the sum you want to compute. Since the denominator goes to zero proportionally to $\epsilon^2$ for small $\epsilon$ (by the small angle approximation for $\sinh$), $f(\epsilon)$ is finite in the limit $\epsilon\to 0$. In fact, with the definition that $f(0)$ is equal to $\lim_{\epsilon\to 0}f(\epsilon)$, $f$ is analytic at $\epsilon=0$. So let's do a Taylor series for $f(\epsilon)$, expanding it in increasing powers of $\epsilon$:
$$
f(\epsilon) = f(0) + \frac{1}{2}\epsilon^2 f''(\epsilon)|_{\epsilon=0} + O(\epsilon^4)
$$
(There are no odd powers since $f$ is even under $\epsilon\to - \epsilon$.)
The value of $f(0)$ is
$$
f(0) = \lim_{\epsilon\to 0} \frac{\epsilon^2}{4\sinh^2(\epsilon\sqrt{x}/2)} = \frac{1}{x}
$$
by the small angle approximation or L'Hopital's rule, whichever you prefer. The second derivative of $f$ is (according to Mathematica)
$$
f''(\epsilon) = \frac{-2+2\epsilon^2 x + (2+\epsilon^2x)\cosh(\epsilon\sqrt{x})-4\epsilon\sqrt{x}\sinh(\epsilon\sqrt{x})}{8\sinh^{4}(\epsilon\sqrt{x}/2)}
$$
This looks like it diverges as $1/\epsilon^4$, but you can check that the numerator is in fact fourth order in $\epsilon$ as well:
$$
-2+2\epsilon^2 x + (2+\epsilon^2x)\cosh(\epsilon\sqrt{x})-4\epsilon\sqrt{x}\sinh(\epsilon\sqrt{x}) = -\frac{x^2}{12}\epsilon^4 + \frac{x^3}{90}\epsilon^6 + O(\epsilon^8)
$$
so the second derivative is finite at $\epsilon=0$ (If it wasn't, I would have been lying when I said $f$ was analytic) and given by
$$
f''(0) = \lim_{\epsilon\to 0} f''(\epsilon) \\
= \lim_{\epsilon\to 0} \frac{-\frac{x^2}{12}\epsilon^4}{8\sinh^{4}(\epsilon\sqrt{x}/2)}\\
=\lim_{\epsilon\to 0} \frac{-\frac{x^2}{12}\epsilon^4}{\frac{1}2 \epsilon^4x^2}\\
=-\frac{1}{6}
$$
So the Taylor series for $f$ is
$$
f(\epsilon) = \frac{1}x - \frac{1}{12}\epsilon^2 + O(\epsilon^4)
$$
Remember that $f(\epsilon)$ was $\epsilon^2$ times your divergent sum. So the expansion in powers of $\epsilon$ of your divergent sum is
$$
\frac{f(\epsilon)}{\epsilon^2} = \frac{1}{\epsilon^2 x} - \frac{1}{12} + O(\epsilon^2)
$$
"Thus" the dimension of spacetime for bosonic strings is $26$.
