# Ramanujan's Incorrect formula

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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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) – Todd Trimble May 28 '11 at 18:18
Unfortunately, your bulleted question has no answer... :) – David Hansen May 29 '11 at 0:31
@Todd: The second sum actually has a $\beta$ which i had actually typed as $\alpha$ – S.C. May 29 '11 at 3:16
@David: Yeah, It's ok. I actually have a solution at the given link :) Was just looking for another method. – S.C. 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.