Sum of a Telescoping Surd Series

Consider the sequence  
\[\frac{1}{\sqrt{2}+\sqrt{3}}, \: \frac{1}{\sqrt{3}+\sqrt{4}}, \frac{1}{\sqrt{4}+\sqrt{5}},..., \: \frac{1}{\sqrt{n}+\sqrt{n+1}}\]
.
We can find the sum of this series if  
\[n\]
  is finite by rationalising the denominator of each time.
\[\frac{1}{\sqrt{k}+\sqrt{k+1}}= \frac{1}{\sqrt{k}+\sqrt{k+1}} \times \frac{\sqrt{k}-\sqrt{k+1}}{\sqrt{k}-\sqrt{k+1}}=\sqrt{k+1}-\sqrt{k}\]
.
The sequence becomes
\[\sqrt{3}-\sqrt{2}, \: \sqrt{4}-\sqrt{3}, \: \sqrt{5}-\sqrt{4},..., \: \sqrt{n+1}-\sqrt{n}\]
.
Now we can add the series.
\[S_n = ( \sqrt{3}-\sqrt{2})+ (\sqrt{4}-\sqrt{3})+( \sqrt{5}-\sqrt{4})+...+( \sqrt{n+1}-\sqrt{n})\]
.
The last term in each bracket cancels with the first term in the next bracket except for the second term in the first bracket second term in the first bracket. The sum of the series is  
\[S_n = -\sqrt{2})+ \sqrt{n+1}=\sqrt{n+1}-\sqrt{2}\]
.

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