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Can anyone see why it is that if $a$ is large, then $$\log (\sum_m\sum_n \exp(-knm/a)))$$ where $k$ is a constant and $n,m$ take values $1,2,3,...$, can be approximated by $${a\pi^2\over 6k }$$? Cheers!

As Gerry suggested, it would probably help if I added more context. Let $$Z= \sum_m\sum_n \exp(-{nm\hbar \omega\over k_B T}))$$ and the free energy is $$A= -k_BT \log Z$$ and for this $Z$, and at large  $T$, $A$ is said to be approximately $$-k_b^2T^2\pi^2\over 6\hbar \omega$$

user61472
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    Source?${}{}{}$ – Gerry Myerson Feb 08 '13 at 11:31
  • @GerryMyerson: it's from some notes I have... – user61472 Feb 08 '13 at 11:38
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    @GerryMyerson: I have added more info. :) – user61472 Feb 08 '13 at 11:57
  • @user61472 Well, at least $\sum_n e^{-knm/a} = 1/(e^{k m/a}-1)$ – Valtteri Feb 08 '13 at 12:13
  • @GerryMyerson: this looks like the OP wants to evaluate something called a partition function in statistical mechanics. – Ron Gordon Feb 08 '13 at 14:48
  • The only thing I can think of is that $b \int_0^{\infty} dx x (e^{b x}-1)^{-1} = \pi^2/(6 b)$. I thought of turning the single sum into an integral as $b \rightarrow \infty$, but I have no idea where the $x$ in the integrand comes from, or the lower limit of 0. It may have something to do with an application of $\log$ to the sum, but I cannot say for sure right now. – Ron Gordon Feb 08 '13 at 15:55

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