# Convergence of $\sum_{n=1}^{+\infty}\tan \left( \frac{\pi}{n}\right )$

I was trying to find the convergence of the following series:

$$\sum_{n=1}^{+\infty}\tan \left( \frac{\pi}{n}\right )$$

The way I did it is using the comparison test using that $$\tan(x) > x$$, so $$\tan \left( \frac{\pi}{n}\right ) > \frac{\pi}{n} > \frac{1}{n}$$ and I concluded that it diverges because of the divergence of $$\sum_{n=1}^{+\infty}\frac{1}{n}$$.

I was wondering if this is correct and if not what did I do wrong and how to do it properly.

• You might want to start at $n=3$.
– J.G.
Sep 25, 2020 at 16:44
• Thats what bothered me the most because at $n = 1$ the value is $0$, and at $\frac{\pi}{2}$ its not defined. What do we do about that, because the task that I was given, $n$ starts from $1$? Sep 25, 2020 at 16:45
• If $\sum_{n=1}a_n$ diverges, then $\sum_{n=k}a_n$ diverges too
– L F
Sep 25, 2020 at 16:47
• Show your instructor the result for $n\ge3$, and mention the case $n=2$. I'm sure they'll understand. Sep 25, 2020 at 16:48
• while you add a finite quantity of points, yes it does. But remember that those poinsts you add must exists in funcion's domain.
– L F
Sep 25, 2020 at 16:52

Note As @J.G propose, you might start from $$n=3$$ otherwise you can't define $$\tan$$