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rmk 1. After some manipulation of the series, if

$f(x):=1 - \frac{1-x}{1+x}\sum_{k\in\mathbb Z} x^{k^2}$

Then

$S:=\frac{1}{2}\int_0^1\frac{f(x)}{x} dx.$

So this confirms Charles Matthews' hint:

$\theta:=\sum_{k\in\mathbb Z} x^{k^2}$ Is the theta series, indeed, linked to the Jacobi theta function. (But I don't know how to proceed then.)

rmk 2. Maybe writing Max Muller's series as a sum over double indices (k,n)

$S:=\sum_{1\le k\le n} \frac{1}{(n^2+2k-1)(n^2+2k)}$

one can find a smart partition of the set of indices that allows an iterate summation. (very few chances of success). For instance changing the order of summation, i.e.

$S:=\sum_{k=1}^\infty \sum_{n=k}^\infty \frac{1}{(n^2+2k-1)(n^2+2k)}$

The inner sum can be treated via partial fraction decomposition and summed as illustrated by David Speyer; I found (with $\Psi:=\Gamma'/\Gamma$)

$\sum_{k=1}^\infty\ \left( \mathrm{Im}\frac{\Psi(k+i\sqrt{2k-1})}{\sqrt{2k-1}}-\mathrm{Im}\frac{\Psi(k+i\sqrt{2k})}{\sqrt{2k}}\right).$

rmk 1. After some manipulation of the series, if

$f(x):=1 - \frac{1-x}{1+x}\sum_{k\in\mathbb Z} x^{k^2}$

Then

$S:=\frac{1}{2}\int_0^1\frac{f(x)}{x} dx.$

So this confirms Charles Matthews' hint:

$\theta:=\sum_{k\in\mathbb Z} x^{k^2}$ Is the theta series, indeed, linked to the Jacobi theta function. (But I don't know how to proceed then.)

rmk 2. Maybe writing Max Muller's series as a sum over double indices (k,n)

$S:=\sum_{1\le k\le n} \frac{1}{(n^2+2k-1)(n^2+2k)}$

one can find a smart partition of the set of indices that allows an iterate summation. (very few chances of success). For instance changing the order of summation, i.e.

$S:=\sum_{k=1}^\infty \sum_{n=k}^\infty \frac{1}{(n^2+2k-1)(n^2+2k)}$

The inner sum can be treated via partial fraction decomposition and summed as illustrated by David Speyer; I found (with $\Psi:=\Gamma'/\Gamma$)

$\sum_{k=1}^\infty\ \left( \mathrm{Im}\frac{\Psi(k+i\sqrt{2k-1})}{\sqrt{2k-1}}-\mathrm{Im}\frac{\Psi(k+i\sqrt{2k})}{\sqrt{2k}}\right)=0.2666107..$

rmk 1. After some manipulation of the series, if

$f(x):=1 - \frac{1-x}{1+x}\sum_{k\in\mathbb Z} x^{k^2}$

Then

$S:=\frac{1}{2}\int_0^1\frac{f(x)}{x} dx.$

So this confirms Charles Matthews' hint:

$\theta:=\sum_{k\in\mathbb Z} x^{k^2}$ Is the theta series, indeed, linked to the Jacobi theta function. (But I don't know how to proceed then.)

rmk 2. Maybe writing Max Muller's series as a sum over double indices (k,n)

$S:=\sum_{1\le k\le n} \frac{1}{(n^2+2k-1)(n^2+2k)}$

one can find a smart partition of the set of indices that allows an iterate summation. (very few chances of success). For instance changing the order of summation, i.e.

$S:=\sum_{k=1}^\infty \sum_{n=k}^\infty \frac{1}{(n^2+2k-1)(n^2+2k)}$

The inner sum can be treated via partial fraction decomposition and summed as illustrated by David Speyer; I found

$\sum_{k=1}^\infty\ \left( \mathrm{Im}\frac{\Psi(k+i\sqrt{2k-1})}{\sqrt{2k-1}}-\mathrm{Im}\frac{\Psi(k+i\sqrt{2k})}{\sqrt{2k}}\right).$

rmk 1. After some manipulation of the series, if

$f(x):=1 - \frac{1-x}{1+x}\sum_{k\in\mathbb Z} x^{k^2}$

Then

$S:=\frac{1}{2}\int_0^1\frac{f(x)}{x} dx.$

So this confirms Charles Matthews' hint:

$\theta:=\sum_{k\in\mathbb Z} x^{k^2}$ Is the theta series, indeed, linked to the Jacobi theta function. (But I don't know how to proceed then.)

rmk 2. Maybe writing Max Muller's series as a sum over double indices (k,n)

$S:=\sum_{1\le k\le n} \frac{1}{(n^2+2k-1)(n^2+2k)}$

one can find a smart partition of the set of indices that allows an iterate summation. (very few chances of success). For instance changing the order of summation, i.e.

$S:=\sum_{k=1}^\infty \sum_{n=k}^\infty \frac{1}{(n^2+2k-1)(n^2+2k)}$

The inner sum can be treated via partial fraction decomposition and summed as illustrated by David Speyer; I found (with $\Psi:=\Gamma'/\Gamma$)

$\sum_{k=1}^\infty\ \left( \mathrm{Im}\frac{\Psi(k+i\sqrt{2k-1})}{\sqrt{2k-1}}-\mathrm{Im}\frac{\Psi(k+i\sqrt{2k})}{\sqrt{2k}}\right).$

rmk 1. After some manipulation of the series, if

$f(x):=1 - \frac{1-x}{1+x}\sum_{k\in\mathbb Z} x^{k^2}$

Then

$S:=\frac{1}{2}\int_0^1\frac{f(x)}{x} dx.$

So this confirms Charles Matthews' hint:

$\theta:=\sum_{k\in\mathbb Z} x^{k^2}$ Is the theta series, indeed, linked to the Jacobi theta function. (But I don't know how to proceed then.)

rmk 2. Maybe writing Max Muller's series as a double sum

$S:=\sum_{1\le k\le n} \frac{1}{(n^2+2k-1)(n^2+2k)}$

one can find a smart partition of the set of indices that allows an iterate summation. (very few chances of success). For instance changing the order of summation, i.e.

$S:=\sum_{k=1}^\infty \sum_{n=k}^\infty \frac{1}{(n^2+2k-1)(n^2+2k)}$

The inner sum can be treated via partial fraction decomposition and summed as illustrated by David Speyer; I found

$\sum_{k=1}^\infty \mathrm{Im}\frac{\Psi(k+i\sqrt{2k-1})}{\sqrt{2k-1}}-\sum_{k=1}^\infty \mathrm{Im}\frac{\Psi(k+i\sqrt{2k})}{\sqrt{2k}}.$

rmk 1. After some manipulation of the series, if

$f(x):=1 - \frac{1-x}{1+x}\sum_{k\in\mathbb Z} x^{k^2}$

Then

$S:=\frac{1}{2}\int_0^1\frac{f(x)}{x} dx.$

So this confirms Charles Matthews' hint:

$\theta:=\sum_{k\in\mathbb Z} x^{k^2}$ Is the theta series, indeed, linked to the Jacobi theta function. (But I don't know how to proceed then.)

rmk 2. Maybe writing Max Muller's series as a sum over double indices (k,n)

$S:=\sum_{1\le k\le n} \frac{1}{(n^2+2k-1)(n^2+2k)}$

one can find a smart partition of the set of indices that allows an iterate summation. (very few chances of success). For instance changing the order of summation, i.e.

$S:=\sum_{k=1}^\infty \sum_{n=k}^\infty \frac{1}{(n^2+2k-1)(n^2+2k)}$

The inner sum can be treated via partial fraction decomposition and summed as illustrated by David Speyer; I found

$\sum_{k=1}^\infty\ \left( \mathrm{Im}\frac{\Psi(k+i\sqrt{2k-1})}{\sqrt{2k-1}}-\mathrm{Im}\frac{\Psi(k+i\sqrt{2k})}{\sqrt{2k}}\right).$