The discussion revolves around evaluating the sum $\displaystyle \sum_{k=1}^{49} \dfrac{1}{\sqrt{ k+\sqrt{k^2-1}}}$. Participants explore different approaches and simplifications related to this mathematical expression.
There is no disagreement noted in the discussion, but the lack of further exploration or challenge suggests a consensus on the proposed method and result.
The discussion does not address potential limitations or assumptions in the mathematical steps taken, nor does it explore alternative methods for evaluating the sum.
Readers interested in mathematical series, telescoping sums, or specific techniques for evaluating sums in calculus may find this discussion relevant.
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:
$$\frac{1}{\sqrt{2}}\left(\sqrt{50}+\sqrt{49}-1\right) = 5+3\sqrt{2} \approx 9.246$$
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ZaidAlyafey said:nice :) .