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$$\sum\limits_{k=1}^n\arctan\frac{ 1 }{ k }=\frac{\pi}{ 2 }$$ Find value of $n$ for which equation is satisfied.

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Do you mean $\tan^{-1}1/n$? –  Jlamprong May 20 at 14:53
You should correct this expression, that does not make sense. It is probably $\sum_{k=1}^n\tan^{-1}\frac1k$. –  Tom-Tom May 20 at 14:59
$\arctan$ and $\tan^{-1}$ both refer to the same function (under any sane interpretation of indices on trig functions, anyway). –  michaelb958 May 20 at 23:32

2 Answers 2

n=3. By drawing this figure, you can easily know enter image description here

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And how do you know it does't hold for $n=15764538$? :) –  Zircht May 20 at 15:05
@ᛥᛥᛥ The picture itself is already explain why it only holds for $n=3$. –  Tunk-Fey May 20 at 15:32
@Tunk-Fey I know. That's why I put the " :) ". –  Zircht May 20 at 15:44
(it's probably not worth fixing, but) shouldn't that be arctan in the picture, not acrtan? –  Tim S. May 20 at 18:18

Let use write $$s_n=\sum_{k=1}^n \arctan\frac1k.$$ The sequence $(s_n)_{n\in\mathbf N}$ is increasing. We have $s_0=0$, $s_1=\frac\pi4$ and $s_2=\frac\pi4+\arctan\frac12$. As $\frac12<1$, $\tan^{-1}\left(\frac12\right)<\frac\pi4$ and $s_2<\frac\pi2$. Let us compute $s_3$ using the arctan addition formula $$s_3=\frac\pi4+\arctan\frac12+\arctan\frac13=\frac\pi4+\arctan\frac{\frac12+\frac13}{1-\frac12\frac13}=\frac\pi4+\arctan1=\frac\pi2.$$ $n=3$ is a solution. As $s_4>s_3$, it's the only one.

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