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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 $-\frac1{12}$?

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\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*}

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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 $26$.

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  • $\begingroup$ Thank you! I'm mostly wondering about where the -1/12 comes from actually. Could you please expand on that? $\endgroup$
    – user46348
    Aug 30 '14 at 18:34
  • $\begingroup$ Sure: see my edit $\endgroup$ Aug 30 '14 at 20:13

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