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I want to know whether the series $\displaystyle{% \sum_{n=1}^{\infty }\left[{\pi \over 2} - \arctan\left(n\right)\right ]}$ converges or not.

Some series such as $\sum_{n=1}^{\infty}\sin \frac1n$, $\sum_{n=1}^{\infty}\tan \frac1n$ are solved by the comparison test with $\sum_{n=1}^{\infty}\frac1n$. But the given series is not compared with $\sum_{n=1}^{\infty}\frac1n$. Is there another way to determine whether the series converges or not?

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marked as duplicate by Jyrki Lahtonen, Davide Giraudo, Amitesh Datta, Old John, Grigory M Dec 31 '13 at 9:47

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Integral test gives you an answer, right? $\endgroup$ – T.J. Gaffney Dec 31 '13 at 7:14
  • $\begingroup$ This may help math.stackexchange.com/questions/308377/… Also, note that $\frac{\pi}{2}-\arctan n = \arctan \frac{1}{n}$. $\endgroup$ – Zhoe Dec 31 '13 at 7:18
  • $\begingroup$ Tangent of the complementary angle is the reciprocal, so $(\pi/2)-\arctan n=\arctan(1/n)$ $\endgroup$ – Jyrki Lahtonen Dec 31 '13 at 7:25
  • $\begingroup$ $\displaystyle{\large{\pi \over 2} - \arctan\left(1 \over n\right) \sim {1 \over n} - {1 \over 3n^{3}}}$ when $\displaystyle{\large n \gg 1}$. $\endgroup$ – Felix Marin Dec 31 '13 at 8:05
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Begin by noting that $$\frac{\pi}{2} - \arctan n = \int_{n}^{\infty} \frac{1}{1 + x^2}$$

Now to estimate the integral, we can use

$$\int_{n}^{\infty} \frac{1}{1 + x^2} \ge \int_n^{\infty} \frac{1}{2}\frac{1}{x^2}$$

for sufficiently large $n$. Compare with the harmonic series.

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    $\begingroup$ Would the downvoter please share the reason? If there's a mistake or correction, I'd be glad to address it. $\endgroup$ – user61527 Dec 31 '13 at 7:18
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    $\begingroup$ I was just going to ask the same question. it seems a very ingenious answer to me! $\endgroup$ – David Holden Dec 31 '13 at 7:20
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    $\begingroup$ Nothing wrong with this. It may be simpler to observe that for $n>0$ we have $(\pi/2)-\arctan n=\arctan(1/n)$. For $x$ close to zero we have $\arctan x\approx x$ then leads to the same conclusion. $\endgroup$ – Jyrki Lahtonen Dec 31 '13 at 7:23
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Personally I like T.Bongers solution best, but one more method you could try is the integral test.

$$\sum_{k=1}^{\infty}\arctan{\frac{1}{k}}\geq\int_1^{\infty}\arctan{\frac{1}{x}}\,dx\\ =\left(x\arctan{\frac{1}{x}}\right)|_1^{\infty}+\int_1^{\infty}\frac{x}{1+x^2}dx\\ =1-\frac{\pi}{4}+\lim_{u\rightarrow\infty}\frac12\log{\frac{u^2+1}{2}}=\infty$$

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As you know, we have $\arctan n+\arctan \frac{1}{n}=\frac{\pi}{2}$. It is sufficient to show that $\sum_{n=1}^{\infty }\left ( \arctan \frac{1}{n}\right )$ is not converges.

Since we have $\;\lim_{n \to \infty} \frac{\arctan(1/n)}{1/n}=1,\;$ by comparison criteria, we can conclude that this series is not converges, as $\displaystyle \sum_{n=1}^\infty \dfrac 1n\;$ is not converges.

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