Average order of $\frac{d(n)}{n^\alpha}$ where $\alpha>0,\alpha\neq 1$. In Apostol's analytic number theory book,the following problem is given as an exercise:
Show that, $\sum\limits_{n\leq x}\frac{d(n)}{n^\alpha}=\frac{x^{1-\alpha}\log x}{1-\alpha}+\zeta^2(\alpha)+O(x^{1-\alpha})$,where $d(n)$ is the divisor function.Can anyone help me solve this?.My attempt is as follows:
$\sum\limits_{n\leq x}\frac{d(n)}{n^\alpha}=\sum\limits_{n\leq x}\sum\limits
_{d|n}\frac{1}{n^\alpha}=\sum\limits_{k\leq x}\sum\limits
_{d\leq x/k}\frac{1}{k^\alpha d^\alpha}$.
But then I expanded $\sum\limits_{d\leq x/k}\frac{1}{d^\alpha}$ using Euler's summation formula.I did the following:
$\sum\limits_{n\leq x}\frac{1}{n^\alpha}=\zeta(\alpha)+\frac{x^{1-\alpha}}{1-\alpha}+O(\frac{1}{x^\alpha})$.
 A: As you stated, the key formula is here
\begin{equation}
\label{eq:key}
\sum_{n\leq x }\frac{1}{n^\alpha} = \zeta(\alpha) + \frac{x^{1-\alpha}}{1-\alpha} + \mathcal{O}\left(\frac{1}{x^\alpha}\right)
\end{equation}
Applying this formula at the inner sum yields
\begin{align*}
\sum_{n\leq x}\frac{d(n)}{n^\alpha} &= \sum_{k\leq x} \sum_{d\leq x/k} \frac{1}{k^\alpha d^\alpha} = \sum_{k \leq x} \frac{1}{k^\alpha}\sum_{d \leq x/k} \frac{1}{d^\alpha} \\
&= \sum_{k \leq x} \frac{1}{k^\alpha}\left(\frac{(x/k)^{1-\alpha}}{1- \alpha} + \zeta(\alpha) + \mathcal{O}\left(\frac{k^\alpha}{x^\alpha}\right)\right).
\end{align*}
Now rearranging the last sum gives
\begin{align*}
\sum_{n\leq x} \frac{d(n)}{n^\alpha} &= \frac{x^{1-\alpha}}{1-\alpha} \sum_{k \leq x}\frac{1}{k} + \zeta(\alpha)\sum_{k\leq x} \frac{1}{k^\alpha} + \mathcal{O}\left(x^{1-\alpha}\right).
\end{align*}
Now if you apply the known formula $\sum_{n\leq x} \frac{1}{n} = \log x + C + \mathcal{O}\left(x^{-1}\right)$ and the key formula once more, you should be able to find the result.
