The evaluation of the sum $\displaystyle \sum_{k=1}^{49} \dfrac{1}{\sqrt{ k+\sqrt{k^2-1}}}$ simplifies to $\frac{1}{\sqrt{2}}\left(\sqrt{50}+\sqrt{49}-1\right)$, which results in the final value of $5 + 3\sqrt{2} \approx 9.246$. The transformation of the original expression involves recognizing the telescoping nature of the sum, allowing for a straightforward calculation. This method effectively utilizes properties of square roots and simplification techniques to arrive at the conclusion.
PREREQUISITESMathematicians, students studying calculus or algebra, and anyone interested in advanced summation techniques and series evaluation.
anemone said:Evaluate $\displaystyle \sum_{k=1}^{49} \dfrac{1}{\sqrt{ k+\sqrt{k^2-1}}}$
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]
ZaidAlyafey said:nice :) .