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What is the value of \(\sum_{n=1}^\infty (\tan^{-1}\sqrt{n}-\tan^{-1}\sqrt{n+1})\)?

 Sep 15, 2019
 #1
avatar+507 
+1

it converges to cot(1)-cot(sqrt(m+1))

 Sep 15, 2019
edited by Davis  Sep 15, 2019
 #2
avatar+507 
+1

tangent is cyclic

so the value of tangent is cyclic

so its converges to invinity

 Sep 15, 2019
 #3
avatar+23135 
+2

What is the value of \(\sum \limits_{n=1}^\infty \left(\tan^{-1}(\sqrt{n})-\tan^{-1}(\sqrt{n+1})\right)\)?

\(I\ assume \ \boxed{\tan^{-1}(x) = \arctan(x)} \)

 

 

\(\begin{array}{|rcll|} \hline && \sum \limits_{n=1}^\infty \left(\arctan(\sqrt{n})-\arctan(\sqrt{n+1})\right) \\\\ &=& \overbrace{ \arctan(\sqrt{1})-\arctan(\sqrt{2}) }^{n=1} \\ && + \quad\qquad \qquad \quad \overbrace{ \arctan(\sqrt{2})-\arctan(\sqrt{3}) }^{n=2} \\ && + \qquad\qquad \qquad \qquad \qquad \quad\quad \overbrace{ \arctan(\sqrt{3})-\arctan(\sqrt{4}) }^{n=3} \\ && \cdots \\ && + \overbrace{ \arctan(\sqrt{\infty})-\arctan(\sqrt{\infty+1}) }^{n=\infty} \\ \hline \end{array} \)

 

\(\text{shorten} \ldots \)

\(\begin{array}{|rcll|} \hline && -\arctan(\sqrt{2})+\arctan(\sqrt{2}) \\ && -\arctan(\sqrt{3})+\arctan(\sqrt{3}) \\ && -\arctan(\sqrt{4})+\arctan(\sqrt{4}) \\ && \cdots \\ && -\arctan(\sqrt{\infty}) + \arctan(\sqrt{\infty}) \\ &=& 0 \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline && \mathbf{\sum \limits_{n=1}^\infty \left(\arctan(\sqrt{n})-\arctan(\sqrt{n+1})\right)} \\\\ &=& \arctan(\sqrt{1})-\arctan(\sqrt{\infty+1}) \\ &=& \arctan(1)-\arctan( \infty ) \\\\ &=& \dfrac{\pi}{4} - \dfrac{\pi}{2} \\\\ &=& -\dfrac{\pi}{4} \\\\ &=& \mathbf{-0.7853981634} \\ \hline \end{array}\)

 

laugh

 Sep 15, 2019

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