Let's define $$ f(x) = \sum\limits_{k=1}^{\infty} \sigma_1(k) x^k, \quad \text{where } \sigma_1(k) = \sum\limits_{d|k}d $$
Is there a way to approach the root $x_0$ of this series, where $-1 < x_0 < 0$ ? I tried to squeeze it and using geometric series, but its too far away.
On
With the help of Mathematica I defined $$f(x)=\sum _{k=1}^{1000} \sigma _1(k) x^k$$ then I applied the Newton's method starting from $x_0=-0.411248$
$$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$
$$ \begin{array}{r|l} n &x_n\\ \hline 0 &-0.411248\\ 1 &-0.4112484791789996\\ 2 &-0.4112484791779548\\ 3 &-0.4112484791779548\\ 4 &-0.4112484791779547\\ 5 &-0.4112484791779548\\ \end{array} $$
bonus
Define Pochhammer symbol $$Poch(q)=\prod _{k=0}^{\infty } \left(1-q^{k+1}\right)$$ The root of $Poch'(q)$ in the interval $(-1,0)$ is the same as the root of $\sum _{k=1}^{\infty} \sigma _1(k) x^k$ in the same interval.
I truncated the sum at $n$ terms and applied three iterations of Newton's Method, starting at $x_0=-0.5$. Here are the results: $$ \left( \begin{array}{cc} 1 & 0 \\ 2 & -0.333384146341463 \\ 3 & 0.0961538461538462 \\ 4 & -0.397485670060330 \\ 5 & -0.500000000000000 \\ 6 & -0.407129807891880 \\ 7 & -0.420277540201558 \\ 8 & -0.409817509129324 \\ 9 & -0.413179097089750 \\ 10 & -0.411220988091159 \\ 11 & -0.411739696601906 \\ 12 & -0.411245203711637 \\ 13 & -0.411344738057402 \\ 14 & -0.411275353266776 \\ 15 & -0.411303377636828 \\ 16 & -0.411288790446980 \\ 17 & -0.411292174341874 \\ 18 & -0.411289268468018 \\ 19 & -0.411289851523367 \\ 20 & -0.411289380413744 \\ 21 & -0.411289515069802 \\ 22 & -0.411289460359527 \\ 23 & -0.411289472717242 \\ 24 & -0.411289463450467 \\ 25 & -0.411289464495063 \\ 26 & -0.411289464563588 \\ 27 & -0.411289464215141 \\ 28 & -0.411289464649207 \\ 29 & -0.411289464488793 \\ 30 & -0.411289464727583 \\ \end{array} \right) $$ Based on this, I would say with high certainty that the root satisfies $x_0\approx -0.41128946$.