# Abs convergent, conditionally convergent or divergent

Classify as absolute convergent, conditionally convergent or divergent:

$$\sum\limits_{n=1}^{\infty} \frac{(-1)^n (\tan^{-1})n}{(n^2+1)}.$$

Answer is absolute convergent, I justify by using p-series. Can someone tell me if its right or wrong to use p-series? My working is $\sum\limits_{n=1}^{\infty}$ $|\frac{(-1)^n (\tan^{-1})n}{(n^2+1)}|$ = $\sum\limits_{n=1}^{\infty}$ $\frac{(\tan^{-1})n}{n^2+1}$

By p-series, it converges. But how do I show it converges absolutely?

• It is not clear what the $n$-th term is. If you mean $(-1)^n\tan^{-1}\left(\frac{n}{n^2+1}\right)$ then we do not have absolute convergence. If you mean $(-1)^n\frac{\tan^{-1}(n)}{n^2+1}$ then we have absolute convergence. Jun 9, 2013 at 15:40
• Sorry for the uncleared text. But the second series is the one. Do you know how show that it is absolute convergence?
– Risa
Jun 9, 2013 at 15:47
• Solution is given below. I suspect you had this solution, but was uncertain of its correctness. Jun 9, 2013 at 15:52
• There's a simpler solution than the one that André Nicolas posted. See my answer below. Jun 9, 2013 at 21:25

We have $0\lt \arctan n \lt \dfrac{\pi}{2}$.
So in absolute value the $n$-th term is $\lt \dfrac{\pi}{2}\dfrac{1}{n^2}$.
By comparison with $\displaystyle\sum_1^\infty \frac{1}{n^2}$, the series converges absolutely.
Another way: We can also use the Integral Test to prove absolute convergence. We want to show that $$\sum_{1}^\infty \frac{\arctan n}{1+n^2}\tag{1}$$ converges. By substitution we can show that $$\int \frac{\arctan x}{1+x^2}\,dx=\frac{1}{2}\left(\arctan x\right)^2+C.$$ Since $$\lim_{x\to\infty} \arctan x=\frac{\pi}{2},$$ it follows that $$\int_1^\infty \frac{\arctan x}{1+x^2}\,dx$$ converges, and therefore by the Integral Test, the series (1) converges.
The answer posted by André Nicolas is good, but this is simpler: $$\sum_{n=0}^\infty \frac{\arctan n}{1+n^2} \le 1 + \frac\pi2\sum_{n=1}^\infty \frac{1}{n^2} < \infty.$$