I actually looked at one of my Questions (posted at MATH.SE) again and found a formula which actually Ramanujan had discovered.

Ramanujan: If $\alpha$ and $\beta$ are positive numbers such that $\alpha \cdot \beta = \pi^{2}$ then, $$\alpha \cdot \sum\limits_{n=1}^{\infty} \frac{n}{e^{2n\alpha} -1} + \beta \cdot\sum\limits_{n=1}^{\infty} \frac{n}{e^{2n\beta}-1} = \frac{\alpha+\beta}{24} -\frac{1}{4}$$

I actually heard that this result is not true. I would like to know where the mistake is and whether something can be rectified in this proof so that, my above problem can be summed by using this result.

  • I would also like to know the Intuitive idea behind discovering such mysterious formulas.
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    $\begingroup$ Is there a typo somewhere? The left side has three instances of $\alpha$ and only one of $\beta$, whereas one expects something more symmetric, by consideration of what happens by interchanging $\alpha$ and $\beta$. (Or, maybe Ramanujan had a typo? :-D) $\endgroup$ – Todd Trimble May 28 '11 at 18:18
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    $\begingroup$ Unfortunately, your bulleted question has no answer... :) $\endgroup$ – David Hansen May 29 '11 at 0:31
  • $\begingroup$ @Todd: The second sum actually has a $\beta$ which i had actually typed as $\alpha$ $\endgroup$ – crskhr May 29 '11 at 3:16
  • $\begingroup$ @David: Yeah, It's ok. I actually have a solution at the given link :) Was just looking for another method. $\endgroup$ – crskhr May 29 '11 at 3:48

I believe this formula is true, provided the $\alpha$ in the second sum is changed to a $\beta$, as suggested by Todd Trimble's comment. Let $$ P(x) = \prod_{n=1}^\infty \frac{1}{1-x^n} $$ be the generating function for the number of partitions of a non-negative integer $n$. Dedekind proved that $P$ satisfies the transformation formula $$ \log P(e^{-2\pi t}) - \log P(e^{-2\pi /t}) = \frac{\pi}{12} \Bigl( \frac{1}{t} - t \Bigr) + \frac{1}{2} \log t $$ for $t > 0$. Differentiating this formula with respect to $t$ gives $$ -\sum_{n=1}^\infty \frac{2\pi n}{e^{2\pi n t}-1} - \frac{1}{t^2} \sum_{n=1}^\infty \frac{2\pi n}{e^{2\pi n/t} -1} = \frac{\pi}{12} \Bigl( -\frac{1}{t^2} - 1\Bigr) + \frac{1}{2t} $$ Now multiply through by $-t/2$ and substitute $\alpha = \pi t$, $\beta = \pi /t$ to get $$ \sum_{n=1}^\infty \frac{\alpha n}{e^{2n\alpha}-1} + \sum_{n=1}^\infty \frac{\beta n}{e^{2n\beta}-1} = \frac{1}{24}(\beta+\alpha) - \frac{1}{4}$$ which is Ramanujan's formula.

The transformation formula for $P$ is related to the theory of modular forms, of which the Eisenstein series mentioned in Derek Jennings' answer to your question on math.stackexchange are important examples. Briefly, if we define $$ \eta(\tau) = \frac{e^{2\pi i \tau/24}}{P(e^{2\pi i \tau})} = e^{2\pi i \tau/24} \prod_{n=1}^\infty (1-e^{2\pi i n \tau}), $$ then $\eta(\tau)^{24}$ is a modular form of weight $12$. As such, $\eta$ satisfies the identity $$ \eta(-1/\tau) = \sqrt{-i \tau}\; \eta(\tau). $$ The transformation formula for $P$ follows by setting $\tau = it$ and taking logs.

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  • $\begingroup$ nice explanation! $\endgroup$ – SGP May 29 '11 at 0:05
  • $\begingroup$ It seem's I have made a mistake. May be Ramanujan had a wrong proof. Thanks for the wonderful proof. $\endgroup$ – crskhr May 29 '11 at 5:16
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    $\begingroup$ Well, to the extent that Ramanujan "proved" things. My understanding is that he didn't much prove things in the modern sense, wasn't much interested in proving things unless Hardy wanted him to. (In a more ancient sense of the word "proof" as meaning "test", he must have had a lot of methods to satisfy himself personally.) $\endgroup$ – Todd Trimble May 29 '11 at 10:39

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