# limit of a sequence; using comparison test

Hey in one of my exercises I've come upon something I can't get the right result. The exercise is to see if a certain sequence converges or diverges - and especially using the comparison test.

The sequence is $$\sum_{n=1}^{\infty}{\frac{\arctan(n)}{n^{1.2}}}$$ Now I know the arctangent approaches $\left. {\pi} \middle/ {2} \right.$. So from the comparison test I can say (correct me if wrong): $$\sum_{n=1}^{\infty}{\frac{\arctan(n)}{n^{1.2}}} \leq \sum_{n=1}^{\infty}{\frac{\left. {\pi} \middle/ {2} \right.}{n^{1.2}}}$$ Which is a p-type polynom ($\frac{1}{n^p}$). with p larger than 1, so the right hand converges. And by the comparison test this also means that the left hand converges (as the equation is always larger than 0).

However the result sheet says it is a divergent series. What did I do wrong?

• It's convergent; and your reasoning is fine (though you should say $\arctan$ approaches $\pi/2$ from below, or something similar, if your inequality is to be fully justified). – David Mitra Jan 12 '14 at 14:08