Average order of the reciprocal of sum of divisors

A standard method in the mean value of multiplicative arithmetic function is the use of Dirichlet series.

Let \begin{align} G(s)&=\prod_p \left(1+\frac{p-1}{p^s(p^2-1)}+\frac{p-1}{p^{2s}(p^3-1)}+\cdots\right)\left(1-\frac1{p^{s+1}}\right)\\&=\prod_p\left(1+\sum_{n=1}^{\infty} \frac{p-1}{p^{ns}(p^{n+1}-1)}\right)\left(1-\frac1{p^{s+1}}\right)\\ &=\prod_p\left(1-\sum_{n=1}^{\infty} \frac{(p-1)^2}{p^{ns}(p^{n+1}-p)(p^{n+1}-1)}\right)=\sum_{n=1}^{\infty} \frac{g(n)}{n^s} \end{align} which converges absolutely for $\Re(s)>-1$.

Then we have $$ \frac1{\sigma(n)}=\sum_{d|n} g(d) \frac nd = \sum_{dk=n} g(d)\frac 1k. $$

Then \begin{align} \sum_{n\leq x}\frac1{\sigma(n)}&=\sum_{d\leq x}g(d)\sum_{k\leq \frac xd} \frac1k\\&=\sum_{d\leq x}g(d)\left( \log\frac xd + \gamma+O\left(\frac dx\right)\right)\\&=\sum_{d\leq x} g(d)\left(\log x - \log d + \gamma\right) + O(x^{-1+\epsilon})\\ &=G(\log x + \gamma)+H+O(x^{-1+\epsilon}), \end{align} where $$ G=G(0), \ H=G'(0). $$ To compute $G(0)$, we put $s=0$ in the Euler product, $$ G(0)=\prod_p \left(1+\sum_{n=1}^{\infty} \frac{p-1}{p^{n+1}-1}\right)\left(1-\frac1p\right). $$ To compute $G'(0)$, we use the logarithmic differentiation of the Euler product, $$ \frac{G'(0)}{G(0)}=\sum_p \left(\frac{-\sum \limits_{n=1}^{\infty} \frac{n(p-1)}{p^{n+1}-1}\log p}{1+\sum\limits_{n=1}^{\infty} \frac{p-1}{p^{n+1}-1}} +\frac{\log p}{p-1}\right). $$ Running the SAGE code,

s=1
for p in primes(1,10000):
    b=0
    for n in range(2,40):
        b+=1/(p^n-1)
    s=s*(1+(p-1)*b)*(1-1/p)
print numerical_approx(s)

gives an approximation of $G$, 0.672744912303011

t=0 
for p in primes(1,10000):
    b=0
    c=0
    for n in range(2,40):
        b+=1/(p^n-1)
        c+=1/(p^n-1)*(n-1)
    t+=(-(p-1)*c)/(1+(p-1)*b)*log(p)+(log(p)/(p-1))
print numerical_approx(t)

gives an approximation of $G'(0)/G(0)$, 0.507239459457213

Then $G\gamma + H$ is approximately

.672744912303011(.577215664901533+.507239459457213)

=0.72956166753