For $n\geq 1$ and $x \gt 0$, define
$ R_n(x)=\frac{1}{n}\sum_{k=1}^{n}\frac{1}{(x+\frac{2k-1}{2n})^2} $
then $R_n(x)$ is a Riemann sum, which converges to the integral
$ R=\int_{0}^1 \frac{dt}{(x+t)^2}=\frac{1}{x}-\frac{1}{x+1} $
We have
$ R-R_1(x)=\frac{1}{x(x+1)(2x+1)^2} $
$ R-R_2(x)=\frac{16x^2+16x+9}{x(x+1)(4x+1)^2(4x+3)^2} $
$ R-R_3(x)=\frac{3888x^4 + 7776x^3 + 6984x^2 + 3096x + 675}{x(x+1)(6x+1)^2(6x+3)^2(6x+5)^2} $
Thus for $n\leq 3$, the numerator of $R-R_n(x)$ has positive coefficients. Can anyone show that $R>R_n(x)$ for any $n$ and $x$, by showing those coefficients are always positive, or by any other method?
By the way, this might look like homework, but it isn't.