# Does the series $\;\sum\limits_{n=0}^{\infty}\left(\frac{\pi}{2} - \arctan(n)\right)$ converge or diverge?

I tried everything that I know but I couldn't solve this series:

$$\sum_{n=0}^{\infty} \left(\frac{\pi}{2} - \arctan(n)\right)$$

Does it diverge or converge?

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Note $\frac{\pi}2-\arctan(n)=\arctan(\frac{1}n)$. Do you have any series approximation for $\arctan(x)$ when $x$ is small? –  Thomas Andrews Feb 19 '13 at 19:48
Given that it's a decreasing function, you could use the integral test for convergence, although Thomas Andrews' method is more elegant. –  Alyosha Feb 19 '13 at 19:55
@ThomasAndrews You don't need series approximations, limit comparison test is enough.... –  N. S. Feb 19 '13 at 19:56
True, you can find limits for $\arctan\frac{1}{n}$ in other ways, @N.S., I just proposed the first that came to mind. –  Thomas Andrews Feb 19 '13 at 19:58
Nice question. This series is divergent, but may have a finite sum if you apply the Euler-MacLaurin formula or the Abel-Plana equation to sum this. (Basically this is called an analytic continuation of some kind). –  Arucard May 31 '13 at 21:55

Hint: $$\sum_{n=0}^{\infty} \left(\frac{\pi}{2} - \arctan(n)\right) = \dfrac{\pi}{2} + \sum_{n=1}^{\infty} \left(\arctan\left(\frac 1n\right)\right)$$

Then clearly, as $\;n \to \infty,\; \dfrac 1n \to 0$

Perhaps use the limit comparison test? Perhaps the integral test?

To use the limit comparison test, as suggested by Mhenni in the comment below, consider $\displaystyle \sum b_n = \sum \left(\dfrac 1n\right)\,,\;\,$ and note that $$\;\lim_{n \to \infty} \frac{\arctan(1/n)}{1/n}=1,\;$$ meaning the two series either converge together or diverge together. So apply what you know about the behavior of $\displaystyle \sum_{n=1}^\infty \dfrac 1n\;$ to the task at hand.

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Why are these two series equal? –  g3d Feb 19 '13 at 19:58
I don't understand how this answer is useful. The series diverges and it falls out pretty quickly by the integration test. –  JSchlather Feb 19 '13 at 19:59
Because $\dfrac{\pi}{2} - \arctan(n) = \arctan\left(\dfrac 1n\right)$ –  amWhy Feb 19 '13 at 20:00
@JacobSchlather It is useful because we know some bounds for $\arctan x$ when $x$ is small. amWhy just hasn't spelled out that step. –  Thomas Andrews Feb 19 '13 at 20:04
You can make the comparison test by considering $b_n=\frac{1}{n}$ and see that $\lim_{n \to \infty} \frac{\arctan(1/n)}{1/n}=1,$ which implies the two series either diverge together or converge together. –  Mhenni Benghorbal Feb 19 '13 at 21:16

The serie $\sum \arctan \frac{1}{n}$ is divergent for two reasons: first, it's a positive serie and second $\arctan\frac{1}{n}\sim \frac{1}{n}$ so by comparaison we conclude since the serie $\sum \frac{1}{n}$ is divergent.

To explain more why the positivity of the serie is important to use comparaison test, this is a counterexample : the serie $\sum \log(1+\frac{(-1)^n}{\sqrt{n}})$ is divergent while $\log(1+\frac{(-1)^n}{\sqrt{n}})\sim\frac{(-1)^n}{\sqrt{n}}$ and the serie $\sum \frac{(-1)^n}{\sqrt{n}}$ is convergent

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