If $d(n) = \sum_{d|n}1$ is the divisor function, it is a fact that $d$ has summatory behaviour
\begin{equation}
\sum_{n \leq x} d(n) = x \log x + (2 \gamma – 1)x + O(\sqrt{x}).
\end{equation}
I'm interested in the behaviour of $\sum_{n \leq x} \frac{d(n)}{n}$ for large $x$. Now, this answer uses the Dirichlet Hyperbola Method, which I understand, to get the result
\begin{equation}
\sum_{n \leq x} \frac{d(n)}{n} = \frac{1}{2} \log^2 x + 2 \gamma \log x + \gamma^2 -2\gamma_1 + O\left(\frac{1}{\sqrt x}\right).
\end{equation}
However, if I try using Abel summation (motivated by this similar question), I get the bound
\begin{equation}
\sum_{n \leq x} \frac{d(n)}{n} = \frac{1}{x}\left(x \log x + (2 \gamma – 1) x + O(\sqrt x)\right) -1 + \int_1^x\frac{t \log t + (2 \gamma – 1) t + O(\sqrt t)}{t^2}dt
\end{equation}
\begin{equation}
=\frac{1}{2}\log^2 x + 2 \gamma\log x + 2(\gamma – 1) + O\left(\frac{1}{\sqrt x} \right).
\end{equation}
Now the numerical constants $\gamma^2-2\gamma_1$ and $2(\gamma-1)$ are not equal; more worryingly I see no way that my approach could have yielded $\gamma_1$ leading me to believe the idea was fundamentally flawed. Would appreciate if someone could aide my understanding.
Best Answer
You probably made the following mistake:
$$\int_1^x \frac{O(\sqrt{t})}{t^2}\,dt = O\left(\frac{1}{\sqrt{x}}\right)$$
That's not right.
Observe that $$\int_1^x \frac{O(\sqrt{t})}{t^2}\,dt = O\left(\int_1^x \frac{\sqrt{t}}{t^2}\,dt\right)=O\left(2-\frac{2}{\sqrt{x}}\right) = O(1)$$
which means that using Abel summation, the best approximation you can get is $$\sum_{n\le x} \frac{d(n)}{n} = \frac{1}{2}\log^2 x + 2 \gamma\log x + O(1)$$