# algebra problem

• Jun 2nd 2009, 10:06 PM
the undertaker
algebra problem
$\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}
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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 PM
pickslides
• Jun 2nd 2009, 11:36 PM
craig
Quote:

Originally Posted by the undertaker
$\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}$

• Jun 2nd 2009, 11:48 PM
pickslides
$\sum_{i=1}^{99} \frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}$

maybe?
• Jun 3rd 2009, 12:01 AM
TheAbstractionist
Quote:

Originally Posted by the undertaker
$\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}$

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

$\sum_{n\,=\,1}^{99}\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\ =\ \frac1{\sqrt1}-\frac1{\sqrt2}\,+\,\frac1{\sqrt2}-\frac1{\sqrt3}\,+\,\frac1{\sqrt3}-\frac1{\sqrt4}\,+\cdots+\,\frac1{\sqrt{99}}-\frac1{\sqrt{100}}$

you can see that all the terms cancel except the first and the last. Hence the sum is $1-\frac1{10}=\frac9{10}.$