<s>[INST] Alternating summation and subtraction of square roots I encountered a problem, to find the integer part of: $$\frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{3} + \sqrt{4}} +...+\frac{1}{\sqrt{99} + \sqrt{100}}$$.
I multiplied the conjugate of each denominator.  Meaning, for $\frac{1}{\sqrt{a} + \sqrt{b}}$, I multiply $\sqrt{a} - \sqrt{b}$.  I get $\sqrt{100} - \sqrt{99} + \sqrt{98} - \sqrt{97} + ... + \sqrt{2} - \sqrt{1}$.  I am stuck here.  How do I simplify this?  Or is it that my method is wrong?
 [/INST] For an approximation, we might start (using a not generally adopted, but not unfamiliar notation) with$$H^{(-1/2)}_n=1 + \sqrt{2} + \ldots + \sqrt{n}=\frac{2}{3}n^{3/2} + \frac{\sqrt{n}}{2} + \zeta\left(-\frac12\right)+o(1),$$ mentioned in this question ($\zeta$ is the famous Riemann zeta function).
Then, $$\sqrt{2n}-\sqrt{2n-1}\pm\ldots=-H^{(-1/2)}_{2n}+2\sqrt{2}\,H^{(-1/2)}_n=\frac12\sqrt{2n}+(2\sqrt{2}-1)\,\zeta\left(-\frac12\right)+o(1).$$
For $n=50$, the main term would be $\displaystyle5+(2\sqrt{2}-1)\,\zeta\left(-\frac12\right)=4.619895\ldots$, while the original sum is $4.6323951\ldots$, as @Henry pointed out, already, $\displaystyle(2\sqrt{2}-1)\,\zeta\left(-\frac12\right)=-0.380105\ldots$ being slightly more accurate. The error (of the order $O(1/n)$) is no surprise.
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