I want to know whether the series $\displaystyle{% \sum_{n=1}^{\infty }\left[{\pi \over 2} - \arctan\left(n\right)\right ]}$ converges or not.
Some series such as $\sum_{n=1}^{\infty}\sin \frac1n$, $\sum_{n=1}^{\infty}\tan \frac1n$ are solved by the comparison test with $\sum_{n=1}^{\infty}\frac1n$. But the given series is not compared with $\sum_{n=1}^{\infty}\frac1n$. Is there another way to determine whether the series converges or not?
Begin by noting that $$\frac{\pi}{2} - \arctan n = \int_{n}^{\infty} \frac{1}{1 + x^2}$$
Now to estimate the integral, we can use
$$\int_{n}^{\infty} \frac{1}{1 + x^2} \ge \int_n^{\infty} \frac{1}{2}\frac{1}{x^2}$$
for sufficiently large $n$. Compare with the harmonic series.
Personally I like T.Bongers solution best, but one more method you could try is the integral test.
$$\sum_{k=1}^{\infty}\arctan{\frac{1}{k}}\geq\int_1^{\infty}\arctan{\frac{1}{x}}\,dx\\ =\left(x\arctan{\frac{1}{x}}\right)|_1^{\infty}+\int_1^{\infty}\frac{x}{1+x^2}dx\\ =1-\frac{\pi}{4}+\lim_{u\rightarrow\infty}\frac12\log{\frac{u^2+1}{2}}=\infty$$
As you know, we have $\arctan n+\arctan \frac{1}{n}=\frac{\pi}{2}$. It is sufficient to show that $\sum_{n=1}^{\infty }\left ( \arctan \frac{1}{n}\right )$ is not converges.
Since we have $\;\lim_{n \to \infty} \frac{\arctan(1/n)}{1/n}=1,\;$ by comparison criteria, we can conclude that this series is not converges, as $\displaystyle \sum_{n=1}^\infty \dfrac 1n\;$ is not converges.
