Nested sums

First note that $\sqrt{n+\sqrt{n^2-1}}=\sqrt{\dfrac{n+1}{2}+2\sqrt{\dfrac{n+1}{2} \cdot \dfrac{n-1}{2}}+\dfrac{n-1}{2}}=\sqrt{\left(\sqrt{\dfrac{n+1}{2}}+\sqrt{\dfrac{n-1}{2}}\right)^2}=\sqrt{\dfrac{n+1}{2}}+\sqrt{\dfrac{n-1}{2}}$ .

So, $\dfrac{1}{\sqrt{n+\sqrt{n^2-1}}}=\dfrac{1}{\sqrt{\dfrac{n+1}{2}}+\sqrt{\dfrac{n-1}{2}}}$ . If we rationalize we get:

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

It's a telescoping sum, so we are left with:

$\dfrac{-\sqrt{1}+\sqrt{50}+\sqrt{49}}{\sqrt{2}}=\dfrac{6+5\sqrt{2}}{\sqrt{2}}=5+3\sqrt{2} \approx \boxed{9.24}$

Apply excel, the answer is 9.24264068711929, not 9.23 though.