= Harmonic divisor sum
Write $B(t)=\sum_{n\leq t}\tau(n)/n$ and $r=\lfloor\sqrt t\rfloor$. The weighted <Dirichlet hyperbola method> gives
$$
B(t)=2\sum_{a\leq r}\frac{H_{\lfloor t/a\rfloor}}a-H_r^2
=\tfrac12\log^2t+2\gamma\log t+c+O\left(\frac{\log(2t)}{\sqrt t}\right).
$$
Indeed, use $H_{\lfloor u\rfloor}=\log u+\gamma+O(1/u)$ and $\sum_{a\leq r}(\log a)/a=\tfrac12\log^2r+c\prime+O(\log(2r)/r)$. 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>.
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