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?
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asked Jan 7, 2017 at 18:21
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$\begingroup$ By pretending arctan is constant for large y, you get an upper bound of the desired form. By pretending log n is constant (OK, less than a fractional power) for large n, you should get a good lower bound. Is there an issue here I am missing? Gerhard "Not Really Seeing The Issue" Paseman, 2017.01.07. $\endgroup$– Gerhard PasemanCommented Jan 7, 2017 at 21:03
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1 Answer
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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.
answered Jan 7, 2017 at 21:12