MHB Evaluating $\displaystyle \sum_{k=1}^{49} \dfrac{1}{\sqrt{ k+\sqrt{k^2-1}}}$

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The evaluation of the sum $\sum_{k=1}^{49} \dfrac{1}{\sqrt{k+\sqrt{k^2-1}}}$ simplifies to $\sum_{k=1}^{49} \sqrt{k-\sqrt{k^2-1}}$. This transformation leads to a telescoping series, resulting in the expression $\frac{1}{\sqrt{2}}(\sqrt{50} + \sqrt{49} - 1)$. The final result of the evaluation is $5 + 3\sqrt{2}$, which approximates to 9.246. The discussion highlights the effectiveness of using algebraic manipulation to simplify complex sums.
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Evaluate $\displaystyle \sum_{k=1}^{49} \dfrac{1}{\sqrt{ k+\sqrt{k^2-1}}}$
 
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anemone said:
Evaluate $\displaystyle \sum_{k=1}^{49} \dfrac{1}{\sqrt{ k+\sqrt{k^2-1}}}$

$$\sum_{k=1}^{49} \frac{1}{\sqrt{ k+\sqrt{k^2-1}}}=\sum_{k=1}^{49} \sqrt{ k-\sqrt{k^2-1}}=\sum_{k=1}^{49} \sqrt{\frac{k+1}{2}+\frac{k-1}{2}-2\sqrt{\frac{k+1}{2}}\sqrt{\frac{k-1}{2}}}$$
$$=\sum_{k=1}^{49} \sqrt{\left(\sqrt{\frac{k+1}{2}}-\sqrt{\frac{k-1}{2}}\right)^2}=\frac{1}{\sqrt{2}}\sum_{k=1}^{49} \sqrt{k+1}-\sqrt{k-1}$$
The sum telescopes and we get:
$$\frac{1}{\sqrt{2}}\left(\sqrt{50}+\sqrt{49}-1\right) = 5+3\sqrt{2} \approx 9.246$$
 
Last edited:
pranav said:
[sp]
$$\sum_{k=1}^{49} \frac{1}{\sqrt{ k+\sqrt{k^2-1}}}=\sum_{k=1}^{49} \sqrt{ k-\sqrt{k^2-1}}=\sum_{k=1}^{49} \sqrt{\frac{k+1}{2}+\frac{k-1}{2}-2\sqrt{\frac{k+1}{2}}\sqrt{\frac{k-1}{2}}}$$
$$=\sum_{k=1}^{49} \sqrt{\left(\sqrt{\frac{k+1}{2}}-\sqrt{\frac{k-1}{2}}\right)^2}=\frac{1}{\sqrt{2}}\sum_{k=1}^{49} \sqrt{k+1}-\sqrt{k-1}$$
the sum telescopes and we get:
$$\frac{1}{\sqrt{2}}\left(\sqrt{50}+\sqrt{49}-1\right) = 5+3\sqrt{2} \approx 9.246$$
[/sp]

nice :) .
 
ZaidAlyafey said:
nice :) .

Thank you! :-)
 
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