Thread: Summing a complex series

Hi, here is a problem involving summing a series involving a complex equation.

Sum this series:

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

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

Please help, BG

Originally Posted by BG5965

Hi, here is a problem involving summing a series involving a complex equation.

Sum this series:

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

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

Please help, BG

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

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

so the sum becomes telescopic.

Sorry, but what does telescopic mean, in terms of maths?

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.

Originally Posted by Danny

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

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

so the sum becomes telescopic.

Did you mean:

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

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

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

Originally Posted by rowe

Did you mean:

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

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

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

I did - thanks for pointing out the typo.

Interestingly, what happens when this becomes an infinite series? That is,

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

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

So does that mean the answer is:

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

I'm a little confused...

Originally Posted by rowe

Interestingly, what happens when this becomes an infinite series? That is,

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

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

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

@BG: $\displaystyle \frac{1}{ \sqrt{1}} - \frac{1}{ \sqrt{1+1}} + $$\displaystyle \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}} = ?$

So its:

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

Also, how did you get from:

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

Thanks, BG