Amazing sum

$\begin{aligned} S & = \sum_{k=2}^{99} \frac 1{(\sqrt 2+k-1)(\sqrt 2+k)(\sqrt 2+k+1)} \\ & = \frac 12 \sum_{k=2}^{99} \left(\frac 1{\sqrt 2+k-1} - \frac 2{\sqrt 2+k} + \frac 1{\sqrt 2+k+1}\right) \\ & = \frac 12 \left(\frac 1{\sqrt 2+1} - \frac 1{\sqrt 2 + 99} - \frac 1{\sqrt 2 + 2} + \frac 1{\sqrt 2 + 100} \right) \\ & \approx \boxed{0.0606} \end{aligned}$

The general term in the sum is of the form

$\dfrac{1}{(x)(x+1)(x+2)}$

which can be rewritten using partial fractions as

$\left( \dfrac{1}{2} \cdot \dfrac{1}{x} \right) - \dfrac{1}{x+1} + \left( \dfrac{1}{2} \cdot \dfrac{1}{x+2} \right)$

Then the sum we're looking for can be rewritten as

$\left[ \left(\frac{1}{2} \frac{1}{\sqrt{2} + 1}\right) - \frac{1}{\sqrt{2} + 2} \color{#3D99F6}{+ \left(\frac{1}{2} \frac{1}{\sqrt{2} + 3}\right)} \right] + \left[ \left(\frac{1}{2} \frac{1}{\sqrt{2} + 2}\right) \color{#3D99F6}{- \frac{1}{\sqrt{2} + 3}} \color{#D61F06}{+ \left(\frac{1}{2} \frac{1}{\sqrt{2} + 4}\right)} \right] + \left[ \color{#3D99F6}{\left(\frac{1}{2} \frac{1}{\sqrt{2} + 3}\right)} \color{#D61F06}{- \frac{1}{\sqrt{2} + 4}} \color{#CEBB00}{+ \left(\frac{1}{2} \frac{1}{\sqrt{2} + 5}\right)} \right] + \left[ \color{#D61F06}{\left(\frac{1}{2} \frac{1}{\sqrt{2} + 4}\right)} \color{#CEBB00}{- \frac{1}{\sqrt{2} + 5}} \color{#333333}{+ \left(\frac{1}{2} \frac{1}{\sqrt{2} + 6}\right)} \right] + \ .... + \\ \\ \left[ \left(\frac{1}{2} \frac{1}{\sqrt{2} + 95}\right) \color{#E81990}{- \frac{1}{\sqrt{2} + 96}} + \color{#20A900}{\left(\frac{1}{2} \frac{1}{\sqrt{2} + 97}\right)} \right] + \left[ \color{#E81990}{\left(\frac{1}{2} \frac{1}{\sqrt{2} + 96}\right)} \color{#20A900}{- \frac{1}{\sqrt{2} + 97}} \color{#EC7300}{+ \left(\frac{1}{2} \frac{1}{\sqrt{2} + 98}\right)} \right] + \left[ \color{#20A900}{\left(\frac{1}{2} \frac{1}{\sqrt{2} + 97}\right)} \color{#EC7300}{- \frac{1}{\sqrt{2} + 98}} \color{#333333}{+ \left(\frac{1}{2} \frac{1}{\sqrt{2} + 99}\right)} \right] + \left[ \color{#EC7300}{\left(\frac{1}{2} \frac{1}{\sqrt{2} + 98}\right)} \color{#333333}{- \frac{1}{\sqrt{2} + 99} + \left(\frac{1}{2} \frac{1}{\sqrt{2} + 100}\right)} \right]$ $$

Note this is a form of a telescoping sum, and that all fractions except for three of the first four and three of the last four will cancel each other out. Thus our sum reduces to

$\begin{aligned} \dfrac{1}{2} \cdot \dfrac{1}{\sqrt{2} + 1} &\color{#D61F06}{- \dfrac{1}{\sqrt{2} + 2} + \dfrac{1}{2} \cdot \dfrac{1}{\sqrt{2} + 2}} + \color{#20A900}{\dfrac{1}{2} \cdot \dfrac{1}{\sqrt{2} + 99} - \dfrac{1}{\sqrt{2} + 99}} \color{#333333}{+ \dfrac{1}{2} \cdot \dfrac{1}{\sqrt{2} + 100}} \\ \\ &= \dfrac{1}{2} \left( \dfrac{1}{\sqrt{2} + 1} - \dfrac{1}{\sqrt{2} + 2} - \dfrac{1}{\sqrt{2} + 99} + \dfrac{1}{\sqrt{2} + 100} \right) \\ \\ & \approx \boxed{0.0606110724} \end{aligned}$