# Summing a complex series

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• July 28th 2009, 06:55 AM
BG5965
Summing a complex series
Hi, here is a problem involving summing a series involving a complex equation.

Sum this series:

$a_1 + a_2 + ... + a_{99}$
where $a_n = \frac{1}{(n + 1)\sqrt{n} + n\sqrt{n + 1}}$ for $n = 1, 2, ..., 99$

I'm not exactly sure how to start this question.

Please help, BG
• July 28th 2009, 07:06 AM
Jester
Quote:

Originally Posted by BG5965
Hi, here is a problem involving summing a series involving a complex equation.

Sum this series:

$a_1 + a_2 + ... + a_{99}$
where $a_n = \frac{1}{(n + 1)\sqrt{n} + n\sqrt{n + 1}}$ for $n = 1, 2, ..., 99$

I'm not exactly sure how to start this question.

Please help, BG

Notice that if you multiply top and bottom by $(n + 1)\sqrt{n} - n\sqrt{n + 1}$ then

$a_n = \frac{(n + 1)\sqrt{n} - n\sqrt{n + 1}}{ (n + 1)^2 n - n^2(n + 1) } = \frac{(n + 1)\sqrt{n} - n\sqrt{n + 1}}{ (n + 1) n } = \frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}}$

so the sum becomes telescopic.
• July 28th 2009, 07:07 AM
BG5965
Sorry, but what does telescopic mean, in terms of maths?
• July 28th 2009, 07:10 AM
Jester
Quote:

Originally Posted by BG5965
Sorry, but what does telescopic mean, in terms of maths?

If you write out the terms you'll see that most of the terms cancel. Only the first and last survive. This is what we mean as a telescopic series.
• July 28th 2009, 01:38 PM
rowe
Quote:

Originally Posted by Danny
Notice that if you multiply top and bottom by $(n + 1)\sqrt{n} - n\sqrt{n + 1}$ then

$a_n = \frac{(n + 1)\sqrt{n} - n\sqrt{n + 1}}{ (n + 1)^2 n - n^2(n + 1)^2 } = \frac{(n + 1)\sqrt{n} - n\sqrt{n + 1}}{ (n + 1) n } = \frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}}$

so the sum becomes telescopic.

Did you mean:

$a_n = \frac{(n + 1)\sqrt{n} - n\sqrt{n + 1}}{ (n + 1)^2 n - n^2(n + 1) }$

in the first part? And if so, the next part makes sense... but how did you get from $\frac{(n + 1)\sqrt{n} - n\sqrt{n + 1}}{ (n + 1) n }$ to $\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}}$...

EDIT: Ah, nm, got it. $\frac{\sqrt{a}}{a} = \frac{1}{\sqrt{a}}$
• July 28th 2009, 02:22 PM
Jester
Quote:

Originally Posted by rowe
Did you mean:

$a_n = \frac{(n + 1)\sqrt{n} - n\sqrt{n + 1}}{ (n + 1)^2 n - n^2(n + 1) }$

in the first part? And if so, the next part makes sense... but how did you get from $\frac{(n + 1)\sqrt{n} - n\sqrt{n + 1}}{ (n + 1) n }$ to $\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}}$...

EDIT: Ah, nm, got it. $\frac{\sqrt{a}}{a} = \frac{1}{\sqrt{a}}$

I did - thanks for pointing out the typo.
• July 29th 2009, 03:21 AM
rowe
Interestingly, what happens when this becomes an infinite series? That is,

$a = \sum_{n=1}^{\infty} \left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)$

I'm guessing it gets closer and closer to 1 since $\lim_{x\to\infty }{{1}\over{\sqrt{x+1}}} = 0$, so a = 1.
• September 1st 2009, 06:29 AM
BG5965
So does that mean the answer is:

$\frac{1}{\sqrt{1}} - \frac{1}{\sqrt{1+1}} + \frac{1}{\sqrt{99}} - \frac{1}{\sqrt{99+1}}
$

I'm a little confused...
• September 1st 2009, 12:01 PM
Defunkt
Quote:

Originally Posted by rowe
Interestingly, what happens when this becomes an infinite series? That is,

$a = \sum_{n=1}^{\infty} \left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)$

I'm guessing it gets closer and closer to 1 since $\lim_{x\to\infty }{{1}\over{\sqrt{x+1}}} = 0$, so a = 1.

Yes, since $S_n = 1-\frac{1}{\sqrt{n+1}}$

@BG: $\frac{1}{ \sqrt{1}} - \frac{1}{ \sqrt{1+1}} +$
$\frac{1}{ \sqrt{1+1}} - \frac{1}{ \sqrt{1+2}} + \frac{1}{ \sqrt{1+2}} - ... - \frac{1}{ \sqrt{99}} + \frac{1}{ \sqrt{99}} - \frac{1}{ \sqrt{1+99}} = ?$

• September 8th 2009, 07:00 AM
BG5965
So its:

$\frac{1}{\sqrt{1}} - \frac{1}{\sqrt{99+1}}$ ?

Also, how did you get from:

$\frac{(n + 1)\sqrt{n} - n\sqrt{n + 1}}{ (n + 1) n }$ to $\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}}$ using $\frac{\sqrt{a}}{a} = \frac{1}{\sqrt{a}}$?

Thanks, BG