Someone in our puzzle group recently posted a question which asked to evaluate the following sum.
$$\frac{1}{e^{\pi}+1}+\frac{3}{e^{3\pi}+1}+\frac{5}{e^{5\pi}+1}+\cdots$$
My first instinct was to check the sum with Wolfram Alpha which showed me that this sum equals $1/24$. This made me curious.
I recast the sum into the following form.
$$\frac{e^{-\pi}}{1+e^{-\pi}}+\frac{3e^{-3\pi}}{1+e^{-3\pi}}+\frac{5e^{-5\pi}}{1+e^{5\pi}}+\cdots$$
The individual terms reminded me of Lambert series which eventually led me into rabbit hole that is to be the content of this post.
Let's define the following function $f(q)$ such that
$\displaystyle f(q)=\frac{q}{1+q}+\frac{3q^3}{1+q^3}+\frac{5q^5}{1+q^5}+\cdots$
Then,
$\displaystyle -f(-q)=\frac{q}{1-q}+\frac{3q^3}{1-q^3}+\frac{5q^5}{1-q^5}+\cdots$
Expanding using the lambert series for divisor functions, we see
$\displaystyle -f(-q)=\sigma_{\text{odd}}(1)q+\sigma_{\text{odd}}(2)q^2+\sigma_{\text{odd}}(3)q^3+\cdots=\sum_n \sigma_{\text{odd}}(n)q^n$
where $\sigma_{\text{odd}}(n)$ where is sum-of-odd-divisors function.
The sum-of-odd-divisors functions is intimately tied to the number of representations of an integer by a sum of four squares (Jacobi four square theorem) and by sum of four triangular numbers (Legendre). See Analogues on two classical theorems on representations of a number and The Parents of Jacobi’s Four Squares Theorem Are Unique, for example.
Therefore,
\displaystyle -f(-q) &= \sum_n \sigma_{\text{odd}}(n)q^n \\ &=\frac{1}{24} \left[8\sum_{n\text{ odd}}\sigma_{\text{odd}}(n)q^n + 24\sum_{n\text{ even}}\sigma_{\text{odd}}(n)q^n + 16\sum_{n\text{ odd}}\sigma_{\text{odd}}(n)q^n \right] \\ &= \frac{1}{24}\left(\sum_n r_4(n)q^n+\sum_n t_4(n) q^n \right) \\ &= \frac{\vartheta_2(q)^4+\vartheta_3(q)^4-1}{24}\end{align}$
$\displaystyle \vartheta_3(q)=\sum_{n=-\infty}^\infty q^{n^2}=\phi(q^2)\prod_{n \text{ odd}}(1+q^n)^2$ and
$\displaystyle \vartheta_2(q)=\sum_{n=-\infty}^\infty q^{(n+1/2)^2}=2\phi(q^2)q^{1/4}\prod_{n \text{ even}}(1+q^n)^2$
where $\displaystyle\phi(q)=\prod_{n}(1-q^n)$ is the Euler function.
Note that
$\displaystyle \prod_{n\text{ odd}}(1-q^n)=\frac{\phi(q)}{\phi(q^2)}$ and $\displaystyle \prod_{n}(1+q^n)=\frac{\phi(q^2)}{\phi(q)}$
Therefore,
$\displaystyle \vartheta_3(-q)=\phi(q^2)\frac{\phi(q)}{\phi(q^2)}\frac{\phi(q)}{\phi(q^2)}$ and $\displaystyle \vartheta_2(-q)=2\phi(q^2)(-q)^{1/4}\frac{\phi(q^4)}{\phi(q^2)}\frac{\phi(q^4)}{\phi(q^2)}$
Now,
$\displaystyle\sqrt\frac{\vartheta_2(-q)}{\vartheta_3(-q)}=\sqrt{2}(-q)^{1/8}\frac{\phi(q^4)}{\phi(q)}$
Putting $q=e^{-\pi}$ in the above expression, we have
$\displaystyle\sqrt\frac{\vartheta_2(-e^{-\pi})}{\vartheta_3(-e^{-\pi})}=\sqrt{2}(-1)^{1/8}e^{-\pi/8}\frac{\phi(e^{-4\pi})}{\phi(e^{-\pi})}$
Using the special values of the Euler function, we see that $\displaystyle \phi(e^{-4\pi})=\frac{e^{\pi/8}}{\sqrt{2}}\phi(e^{-\pi})$
Then,
$\displaystyle\sqrt\frac{\vartheta_2(-e^{-\pi})}{\vartheta_3(-e^{-\pi})}=(-1)^{1/8} \implies \vartheta_2(-e^{-\pi})^4+\vartheta_3(-e^{-\pi})^4=0$
Returning to our expression for $f(q)$, we see that $24f(e^{-\pi})=1-\vartheta_2(-e^{-\pi})^4-\vartheta_3(-e^{-\pi})^4=1$
As $f(e^{-\pi})$ is essentially the sum that we started with, we finally see that
$$\displaystyle\frac{1}{e^{\pi}+1}+\frac{3}{e^{3\pi}+1}+\frac{5}{e^{5\pi}+1}+\cdots=\frac{1}{24}$$
Phew!
As an aside, we could probably arrive the result a little quicker by using some results like
$\displaystyle\lambda(-1+i)=\frac{\vartheta_2(e^{(-1+i)\pi})^4}{\vartheta_3(e^{(-1+i)\pi})^4}=\frac{\vartheta_2(-e^{-\pi})^4}{\vartheta_3(-e^{-\pi})^4}=-1$ (or) $G_1=1\text{ and }g_1=2^{-1/8}$
where $\lambda(n)$ is the Modular lambda function and $g\text{ and }G$ are the Ramanujan g- and G- functions. But I didn't want to bring in additional exotic functions into the problem especially when I can't find a source for the above values of these functions.
Hope you enjoyed this. See ya soon.
Until then, Yours Aye
Me
