Harmonic divisor sum

Write $B(t)={\sum }_{n\le t}\tau (n)/n$ and $r=\lfloor \sqrt{t}\rfloor$. The weighted Dirichlet hyperbola method givesIndeed, use $H_{\lfloor u\rfloor }=\log u+\gamma +O(1/u)$ and ${\sum }_{a\le r}(\log a)/a=\frac{1}{2}{\log }^{2}r+c\prime +O(\log (2r)/r)$. The latter follows by comparing each summand with the integral on its unit interval: the derivative of $(\log t)/t$ has an integrable tail. Since $r=\sqrt{t}+O(1)$, substitution yields the expansion. Here $H_n$ is the harmonic number and $\gamma$ the Euler--Mascheroni constant.