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! :-)

 

[CellCoree]

I would first clarify the context of this summation. Is it part of a larger mathematical equation or problem? What are the units of the summation? These details are important in understanding the significance of the summation.

Next, I would approach the evaluation of this summation by breaking it down into smaller parts. The expression inside the square root, $k^2-1$, can be simplified to $(k-1)(k+1)$. This allows us to rewrite the original summation as $\displaystyle \sum_{k=1}^{49} \dfrac{1}{\sqrt{ k+\sqrt{(k-1)(k+1)}}}$.

We can then further simplify the expression by using the identity $\sqrt{ab} = \sqrt{a} \sqrt{b}$. This gives us $\displaystyle \sum_{k=1}^{49} \dfrac{1}{\sqrt{ k+\sqrt{k-1} \sqrt{k+1}}}$.

Next, we can use the fact that $\sqrt{k-1} \sqrt{k+1} = \sqrt{(k-1)(k+1)} = \sqrt{k^2-1}$. This allows us to rewrite the summation as $\displaystyle \sum_{k=1}^{49} \dfrac{1}{\sqrt{ k+\sqrt{k^2-1}}}$.

At this point, we have simplified the expression as much as possible and can now evaluate the summation. This can be done by plugging in values for $k$ from 1 to 49 and adding up the resulting fractions. Depending on the desired level of precision, this can be done manually or with the help of a calculator or computer program.

In conclusion, as a scientist, I would evaluate this summation by breaking it down into smaller parts and simplifying it before plugging in values and adding them up. The resulting value would depend on the context and units of the summation and could be used to further analyze the problem at hand.