convergence of series, exercise I must prove convergence or divergence of the following series:
$$
\sum_{n=1}^{\infty}
\left[\arctan\left(1 \over n\right) - {n \over n^{2} + 1}\right]
$$
I've tried the integral test:
$$
\int_{1}^{\infty}x\,\arctan\left(1 \over x\right)\,{\rm d}x$$ and this is $<1$ so I would conclude convergence.
However this integral test requires a monotone decreasing function. Could someone give me some pointers to how to prove this?

Hint. 
$$(\arctan t)'=t-\frac{t^3}{3}+\frac{t^5}{5}+\cdots$$
thus
$$
\left|\arctan\frac{1}{n}-\frac{1}{n}\right|\le \frac{1}{3n^3}+\frac{1}{5n^5}+\cdots\le
\frac{1}{3n^3}\left(1+\frac{1}{n^2}+\cdots\right)=\frac{1}{3n^3}\frac{n^2}{n^2-1}\le\frac{1}{n^3},
$$
for $n\ge 2$, and
$$
\left|\,\frac{n}{n^2+1}-\frac{1}{n}\,\right|=\left|\,\frac{n^2}{n(n^2+1)}-\frac{n^2+1}{n(n^2+1)}\,\right|\le \frac{1}{n^3}.
$$
Thus
$$
\left|\,\arctan\frac{1}{n}-\frac{n}{n^2+1}
\,\right|\le\frac{2}{n^3},
$$
for $n\ge 2$.
