I am finding the following first order estimate.



> 
> **Question.** As $y\rightarrow\infty$,
>  $$\sum\_{n=1}^{\infty}\frac{\log n}n\,\arctan\frac{y}n\,\,
> \sim\,\,\frac{\pi}4\log^2y.$$
>  Is it true?
> 
> 
> 



asymptotic estimate for log-tan sum
Yes, it is true.


Your sum has the same asymptotics as the integral $\int\_1^\infty \frac{\log x}{x}\arctan{\frac{y}x}dx$ by standard arguments (the integrated function is decreasing for $x>e$, say, and each specific summand is bounded, this is quite enough).


Next, we denote $x=y/z$ to get the integral $\int\_0^{y}\frac{\log(y)-\log(z)}z\arctan z dz$. We have $\int\_0^{y}\frac{1}z\arctan z dz\sim \frac{\pi}2\log y$ and $\int\_0^{y}\frac{\log(z)}z\arctan z dz\sim \frac{\pi}4\log^2 y$, both by L'Hôpital's rule.


