I have had difficulty with applying Euler's Summation Formula for the following problem:
Using Euler's Summation Formula, prove that for $x\gt1$ and for some constant $C$ such that $|{C}|\lt 1$ $$\sum_{n\leq x} \frac{1}{1+n^2} = \tan^{-1}(x)+C+\mathcal{0}(\frac{1}{1+x^2})$$
I am using Tom Apostol's Analytic Number Theory and have had trouble even starting in the right place with this problem.
I start with $$ \sum_{y < n \leq x}f(n) = \int_y^x f(t)dt + \int_y^x(t-[t])f'(t)dt + f(x)([x] - x) - f(y)([y] - y) $$ I have tried, seemingly incorrectly, to show $$\int \tan^{-1}(x)dx = \frac{1}{1+x^2}+C$$ as a place to start. Can someone help with an approach to this or a hint?