# series divergent or convergent

Does the series $\sum_{n=1}^{\infty} \arctan(13n)$ converge?

I have tried in this way: since $2 \tan^{-1} x=\tan^{-1} \frac{2x}{1-x^{2}}$, $$\sum_{n=1}^{\infty} \arctan(13n)=\sum_{n=1}^{\infty}\frac{1}{2} \tan^{-1} \frac{26n}{1-169n^{2}}.$$ But now I can't proceed further. Please help me out.

• Always try applying the Divergence Test first, i.e. if the limit is not zero, the series diverges. If the limit is zero, other tests must be applied (I usually follow up with the Ratio Test). – electronpusher Apr 2 '17 at 18:20

By the divergence test, $$\lim_{n\to\infty}a_n=\lim_{n\to\infty}\arctan(13n)=\frac{\pi}{2}\neq0$$ Therefore, the series diverges.
Hint: What is $\arctan(13n)$ for $n\to \infty$? Remember that a necessary condition is that $\arctan(13n) \to 0$ for $n \to\infty$.