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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.

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    $\begingroup$ You might want to start at $n=3$. $\endgroup$
    – J.G.
    Sep 25, 2020 at 16:44
  • $\begingroup$ 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 $1$? $\endgroup$
    – Dzamba
    Sep 25, 2020 at 16:45
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    $\begingroup$ If $\sum_{n=1}a_n$ diverges, then $\sum_{n=k}a_n$ diverges too $\endgroup$
    – L F
    Sep 25, 2020 at 16:47
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    $\begingroup$ Show your instructor the result for $n\ge3$, and mention the case $n=2$. I'm sure they'll understand. $\endgroup$
    – player3236
    Sep 25, 2020 at 16:48
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    $\begingroup$ while you add a finite quantity of points, yes it does. But remember that those poinsts you add must exists in funcion's domain. $\endgroup$
    – L F
    Sep 25, 2020 at 16:52

1 Answer 1

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Yes, this is correct, actually, your approach is the correct, fastest one. By comparison test (as you proposed), this series is not convergent since harmonic series is not convergent.

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

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