Originally Posted by
the undertaker $\displaystyle \frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}$
is there a quick way to find out wha the sum of the series would be
if u reolace the n with 1, then 2, then 3 up to.... 99?
If you write out the sum, you have
$\displaystyle \sum_{n\,=\,1}^{99}\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\ =\ \frac1{\sqrt1}-\frac1{\sqrt2}\,+\,\frac1{\sqrt2}-\frac1{\sqrt3}\,+\,\frac1{\sqrt3}-\frac1{\sqrt4}\,+\cdots+\,\frac1{\sqrt{99}}-\frac1{\sqrt{100}}$
you can see that all the terms cancel except the first and the last. Hence the sum is $\displaystyle 1-\frac1{10}=\frac9{10}.$