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Compute \(\sum_{n = 1}^{9800} \frac{1}{\sqrt{n + \sqrt{n^2 - 1}}}\)

 Feb 25, 2019
 #1
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sumfor(n, 1, 9800, (1/(sqrt(n+sqrt(n^2 - 1)))) = 139.2964645562 ...........

 Feb 26, 2019
 #2
avatar+22172 
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Compute 

 

\(\mathbf{\huge{\sum \limits_{n = 1}^{9800} \dfrac{1}{\sqrt{n+\sqrt{n^2-1}} } } } \\\)

 

\(\begin{array}{|rcll|} \hline && \mathbf{\large{\sum \limits_{n = 1}^{9800} \dfrac{1}{\sqrt{n+\sqrt{n^2-1}} } }} \quad | \quad \sqrt{n+\sqrt{n^2-1}}\sqrt{n-\sqrt{n^2-1}}=1 \\\\ &=& \sum \limits_{n = 1}^{9800} \sqrt{n-\sqrt{n^2-1}} \\\\ &=& \sum \limits_{n = 1}^{9800} \sqrt{\dfrac{1}{2}\left( 2n-2\sqrt{n^2-1} \right) } \\\\ &=& \dfrac{1}{\sqrt{2}}\cdot \sum \limits_{n = 1}^{9800} \sqrt{ 2n-2\sqrt{n^2-1} } \quad | \quad \dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2} \\\\ &=& \dfrac{\sqrt{2}}{2}\cdot \sum \limits_{n = 1}^{9800} \sqrt{ 2n-2\sqrt{n^2-1} } \quad | \quad n^2-1 = (n+1)(n-1) \\\\ &=& \dfrac{\sqrt{2}}{2}\cdot \sum \limits_{n = 1}^{9800} \sqrt{ 2n-2\sqrt{n+1}\sqrt{n-1} } \quad | \quad 2n-2\sqrt{n+1}\sqrt{n-1} = \left( \sqrt{n+1}-\sqrt{n-1} \right)^2 \\\\ &=& \dfrac{\sqrt{2}}{2}\cdot \sum \limits_{n = 1}^{9800} \sqrt{ \Big( \sqrt{n+1}-\sqrt{n-1} \Big)^2 } \\\\ &=& \dfrac{\sqrt{2}}{2}\cdot \sum \limits_{n = 1}^{9800} \Big( \sqrt{n+1}-\sqrt{n-1} \Big) \\\\ &=& \dfrac{\sqrt{2}}{2}\cdot \left[~ \sum \limits_{n = 1}^{9800} \Big( \sqrt{n+1} \Big)-\sum \limits_{n = 1}^{9800} \Big( \sqrt{n-1} \Big) ~\right] \quad | \quad \sum \limits_{n = 1}^{9800} \Big( \sqrt{n-1} \Big) = 0+1+ \sum \limits_{n = 3}^{9800} \Big( \sqrt{n-1} \Big)\\\\ &=& \dfrac{\sqrt{2}}{2}\cdot \left[~ \sum \limits_{n = 1}^{9800} \Big( \sqrt{n+1} \Big) -\left(0+1+\sum \limits_{n = 3}^{9800} \Big( \sqrt{n-1} \Big) \right) ~\right] \\\\ &=& \dfrac{\sqrt{2}}{2}\cdot \left[~ \sum \limits_{n = 1}^{9800} \Big( \sqrt{n+1} \Big) - \sum \limits_{n = 3}^{9800} \Big( \sqrt{n-1} \Big)-1 ~\right] \quad | \quad \sum \limits_{n = 3}^{9800} \Big( \sqrt{n-1} \Big) = \sum \limits_{n = 2}^{9799} \Big( \sqrt{n} \Big) \\\\ &=& \dfrac{\sqrt{2}}{2}\cdot \left[~ \sum \limits_{n = 1}^{9800} \Big( \sqrt{n+1} \Big) - \sum \limits_{n = 2}^{9799} \Big( \sqrt{n} \Big)-1 ~\right] \quad | \quad \sum \limits_{n = 1}^{9800} \Big( \sqrt{n+1} \Big) = \sum \limits_{n = 2}^{9801} \Big( \sqrt{n} \Big) \\\\ &=& \dfrac{\sqrt{2}}{2}\cdot \left[~ \sum \limits_{n = 2}^{9801} \Big( \sqrt{n} \Big) - \sum \limits_{n = 2}^{9799} \Big( \sqrt{n} \Big)-1 ~\right] \quad | \quad \sum \limits_{n = 2}^{9801} \Big( \sqrt{n} \Big) = \sum \limits_{n = 2}^{9799} \Big( \sqrt{n} \Big) +\sqrt{9800} +\sqrt{9801} \\\\ &=& \dfrac{\sqrt{2}}{2}\cdot \left[~ \sum \limits_{n = 2}^{9799} \Big( \sqrt{n} \Big) +\sqrt{9800} +\sqrt{9801} - \sum \limits_{n = 2}^{9799} \Big( \sqrt{n} \Big)-1 ~\right] \\\\ &=& \dfrac{\sqrt{2}}{2}\cdot \left[~ \underbrace{\sum \limits_{n = 2}^{9799} \Big( \sqrt{n} \Big)- \sum \limits_{n = 2}^{9799} \Big( \sqrt{n} \Big)}_{=0} +\sqrt{9800} +\sqrt{9801} -1 ~\right] \\\\ &\mathbf{=}& \mathbf{\dfrac{\sqrt{2}}{2}\cdot \left[~ \sqrt{9800} +\sqrt{9801} -1 ~\right]} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{\large{\sum \limits_{n = 1}^{9800} \dfrac{1}{\sqrt{n+\sqrt{n^2-1}} } }} &\mathbf{=}& \mathbf{\dfrac{\sqrt{2}}{2}\cdot \left[~ \sqrt{9800} +\sqrt{9801} -1 ~\right]} \\ &=& \dfrac{\sqrt{2}}{2}\cdot \left[~ 98.9949493661 + 99 -1 ~\right] \\ &=& \dfrac{\sqrt{2}}{2}\cdot 196.994949366 \\ &\mathbf{=}& \mathbf{139.296464556} \\ \hline \end{array} \)

 

laugh

 Feb 27, 2019

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