Let $d(n)$ be the divisor function, that is, $$d(n)=\sum_{1\leq k\leq n, \,k|n} 1.$$ Or equivalently, $d(n)$ can be defined as the coefficients of $\zeta^2(s)$: $$\zeta^2(s)=\sum_{n\geq1} d(n)n^{-s}.$$ Follow this equivalent definition, one can define $d_{\alpha}(n)$ for all $\alpha\geq 1$ by $$\zeta^{\alpha}(s)=\sum_{n\geq 1} d_{\alpha}(n)n^{-s}.$$ In some papers, $d_\alpha (n)$ are called the general divisor functions.
Question: Find the asymptotic on $$\sum_{1\leq n \leq x}\frac{1}{d_{\alpha}(n)}.$$
The following asymptotic formula for case $\alpha=2$(the usual divisor function) was announced by Ramanujan in 1916, and proved by Wilson in 1922:$$\sum_{1\leq n\leq x} \frac{1}{d(n)} \sim A\frac{x}{(\log x)^{1/2}}$$ where $A=\frac{1}{\sqrt{\pi}}\Pi_p \sqrt{p(p-1)}\log\frac{p}{p-1}$. I find some references concerning such asymptotic of general divisor functions but unfortunately, none of them gives the asymptotic formula for $\sum_{n\leq x} 1/d_\alpha(n)$. I guess (of course may false) the result is $$\sum_{1\leq n \leq x} \frac{1}{d_\alpha(n)}\sim C \frac{x}{(\log x)^{1-1/\alpha}},$$ for some suitable constant C depends on $\alpha$.