I was solving simple series involving the function $\arctan(x)$ of the form :
$$\sum \arctan\left(\frac{d}{ax^2+bx+c}\right)$$
where $a,d,b,c$ are constants. This series can be manipulated too by simple substitutions. The basic idea of solving these types was to convert the original sum to a telescoping series by using the relation (for $x,y>0$):
$$\arctan\left(\frac{x-y}{1+xy}\right)=\arctan(x) -\arctan(y)$$
But , for my curiousity , I tried to see if it's possible to evaluate this for higher degree polynomials instead of quadratics. Using online calculators, I observed that :
$$I_{100}ā0.785$$
$$I_{1000}ā0.78538$$
where
$$I_p=\sum_{nā„1}\arctan\left(\frac{1}{n^p}\right)$$
Question:
Can we prove if $$\lim_{p\to\infty}I_p=\frac{\pi}{4}$$ because $$\frac{\pi}{4}ā0.785398$$ which is numerically very close to the sum above ? My knowledge about advanced techniques of calculus to find such sums is very low.
Best Answer
$$ \lim_{p\to\infty}\sum_{n=1}^\infty\arctan\frac1{n^p}=\arctan{1}+\lim_{p\to\infty}\sum_{n=2}^\infty\arctan\frac1{n^p}=\frac\pi4+0. $$
The relation $$ \lim_{p\to\infty}\sum_{n=2}^\infty\arctan\frac1{n^p}=0 $$ follows from summing the inequalities $$0<\arctan\frac1{n^p}<\frac1{n^p}$$ and applying squeeze theorem.