$\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?

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- Jun 2nd 2009, 10:06 PMthe undertakeralgebra problem
$\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?

- Jun 2nd 2009, 11:30 PMpickslides
Your equation is not shown.

- Jun 2nd 2009, 11:36 PMcraig
- Jun 2nd 2009, 11:48 PMpickslides
$\displaystyle \sum_{i=1}^{99} \frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}$

maybe? - Jun 3rd 2009, 12:01 AMTheAbstractionist
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 $\displaystyle 1-\frac1{10}=\frac9{10}.$