How to find $x$ in $\sum\limits_{i=1}^{\infty}\sum\limits_{j=1}^{\infty}x^{ij}=\frac{1}{1-\gamma}$ if $0\lt x\lt1$?

Question

I calculated the approximate sums and it seemed to be the case that $x=\gamma$ so far I cannot prove it.

Answer 1

The Lambert series $$f(x)=\sum_{i,j\geq 1}x^{ij}=\sum_{n\geq 1} d(n) x^n \qquad (x\in(-1,1))\tag{1}$$ has not a simple closed form, but the Dirichlet's hyperbola method ensures $$ \sum_{n=1}^{N}d(n) = N\log N+(2\gamma-1)N+O(\sqrt{N}) \tag{2}$$ leading to the approximation $$ f(x) \approx -\frac{\log(1-x)}{1-x}+\frac{x^2}{2(1-x)}\qquad \forall x\in[0,1).\tag{3}$$

Answer 2

$$\sum\limits_{i=1}^{\infty}\sum\limits_{j=1}^{\infty}x^{ij}=\sum\limits_{i=1}^{\infty}\frac{x^i}{1-x^i}.$$

I don't think there is a closed-form expression (it takes the digamma function). For small $x$, you can approximate with

$$\frac x{1-x}.$$

Anyway, your conjecture seems grossly wrong. By numerical computation, the sum is

$$S(\gamma)\approx2.385957$$

while

$$\frac1{1-\gamma}\approx2.365272$$