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$$\sum\limits_{k=3}^n\arctan\frac{ 1 }{ k }=\frac{\pi}{ 4 }$$ Find value of $n$ for which equation is satisfied.

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If the summation starts at $k=3$, I have the feeling that there no value of $n$ such that $$f(n)=\sum\limits_{k=3}^n\arctan\frac{ 1 }{ k }-\frac{\pi}{ 4 }$$could be equal to $0$.

$f(n)$ increases with $n$ and you have $$f(3)\simeq -0.463648$$ $$f(4) \simeq -0.218669$$ $$f(5) \simeq -0.021273$$ $$f(6) \simeq +0.143875$$.

If the summation would have started at $k=2$, using Fabien's suggestion, the solution would have been $n=3$.

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  • $\begingroup$ Answer given is 47 ... $\endgroup$ – avz2611 May 21 '14 at 6:54
  • $\begingroup$ @user142634 That answer is not correct. $\endgroup$ – 6005 May 21 '14 at 6:57
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$$ \sum\limits_{k=3}^5\arctan\frac{ 1 }{ k }\approx \frac{\pi}{ 4 } $$ To prove that, use : $$ \tan(a+b)=\frac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)} $$

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  • $\begingroup$ The suggestion is good but the statement is untrue. $\endgroup$ – 6005 May 21 '14 at 6:45
  • $\begingroup$ see WolframAlpha and also Claude Leibovici's answer. $\endgroup$ – 6005 May 21 '14 at 6:50

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