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

[anemone]

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

 

[Saitama]

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

Spoiler

$$\sum_{k=1}^{49} \frac{1}{\sqrt{ k+\sqrt{k^2-1}}}=\sum_{k=1}^{49} \sqrt{ k-\sqrt{k^2-1}}=\sum_{k=1}^{49} \sqrt{\frac{k+1}{2}+\frac{k-1}{2}-2\sqrt{\frac{k+1}{2}}\sqrt{\frac{k-1}{2}}}$$
$$=\sum_{k=1}^{49} \sqrt{\left(\sqrt{\frac{k+1}{2}}-\sqrt{\frac{k-1}{2}}\right)^2}=\frac{1}{\sqrt{2}}\sum_{k=1}^{49} \sqrt{k+1}-\sqrt{k-1}$$
The sum telescopes and we get:
$$\frac{1}{\sqrt{2}}\left(\sqrt{50}+\sqrt{49}-1\right) = 5+3\sqrt{2} \approx 9.246$$

 

[alyafey22]

[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 :) .

 

[Saitama]

nice :) .

Thank you! :-)