# Prove the value of the infinite series.

• Jun 19th 2008, 04:08 PM
mathwizard
Prove the value of the infinite series.
Prove that

$\displaystyle 1 + \frac{3}{4} + \frac{3*5}{4*8} + \frac{3*5*7}{4*8*12} + \frac{3*5*7*9}{4*8*12*16} + ... = \sqrt{8}$

Also find the sigma (summation) representation of the series.

EDIT: Can you prove it without using calculus? Thanks.
• Jun 19th 2008, 06:54 PM
Krizalid
The series can be rewritten as $\displaystyle \sum\limits_{n\,=\,0}^{\infty }{\frac{(2n+1)!}{2^{3n}(n!)^{2}}}.$

By using $\displaystyle \frac1{2^{2n}}\binom{2n}n=\frac2\pi\int_0^{\pi/2}\sin^{2n}(x)\,dx$ and after some straightforward calculations we get that the series equals $\displaystyle \frac{4}{\pi }\int_{0}^{\pi /2}{\frac{2+\sin ^{2}x}{\left( 2-\sin ^{2}x \right)^{2}}\,dx},$ which gives the desired result. $\displaystyle \blacksquare$
• Jun 19th 2008, 07:49 PM
ThePerfectHacker
Here is another solution.

Begin be noting as Krizalid said that this is $\displaystyle \sum_{n=0}^{\infty} \frac{(2n+1)}{2^{3n}} {{2n}\choose n} = 2\sum_{n=0}^{\infty}\frac{n}{2^{3n}}{{2n}\choose n} + \sum_{n=0}^{\infty} \frac{1}{2^{3n}} {{2n}\choose n}$.

We will use fact that $\displaystyle \sum_{n=0}^{\infty}{{2n}\choose n} x^n = \frac{1}{\sqrt{1-4x}}$ if $\displaystyle |x| < \frac{1}{4}$.

If we differenciate this term-by-term we get $\displaystyle \sum_{n=1}^{\infty} n{{2n}\choose n}x^{n-1} = 2(1-4x)^{-3/2}$.
Multiply though by $\displaystyle x$ to get, $\displaystyle \sum_{n=0}^{\infty} n {{2n}\choose n}x^n = 2x(1-4x)^{-3/2}$.

Thus, $\displaystyle \sum_{n=0}^{\infty} \frac{n}{2^{3n}} {{2n}\choose n} = 2\cdot \tfrac{1}{8} \cdot (1-4\cdot \tfrac{1}{8} )^{-3/2} = \tfrac{1}{4}\sqrt{8}$

And $\displaystyle \sum_{n=0}^{\infty} \tfrac{1}{2^{3n}}{{2n}\choose n} = (1-4\cdot \tfrac{1}{8} )^{-1/2} = \sqrt{2}$

That means the answer is $\displaystyle \tfrac{1}{2} \sqrt{8} + \sqrt{2} = \sqrt{2}+\sqrt{2} = 2\sqrt{2} = \sqrt{8}$.
• Jun 20th 2008, 06:08 AM
Krizalid
Quote:

Originally Posted by mathwizard

EDIT: Can you prove it without using calculus? Thanks.

Can you stop duplicating questions in others forums, moreover if they're already solved?

I just saw in another forum you posted this same question at 23:46. Did you checked our solutions? Then after you duplicated the post you're askin' "Can you prove it without using calculus? Thanks."

If you want more solutions, ask it right here.

Quote:

Originally Posted by ThePerfectHacker

Here is another solution.

Begin be noting as Krizalid said that this is $\displaystyle \sum_{n=0}^{\infty} \frac{(2n+1)}{2^{3n}} {{2n}\choose n} = 2\sum_{n=0}^{\infty}\frac{n}{2^{3n}}{{2n}\choose n} + \sum_{n=0}^{\infty} \frac{1}{2^{3n}} {{2n}\choose n}$.

Yeah, I have to say that was the hardest part of solving this, it took me one hour to get the general term! Well but I success in solving the problem. :D
• Jun 21st 2008, 03:12 AM
galactus
Quote:

Originally Posted by Krizalid
The series can be rewritten as $\displaystyle \sum\limits_{n\,=\,0}^{\infty }{\frac{(2n+1)!}{2^{3n}(n!)^{2}}}.$

By using $\displaystyle \frac1{2^{2n}}\binom{2n}n=\frac2\pi\int_0^{\pi/2}\sin^{2n}(x)\,dx$ and after some straightforward calculations we get that the series equals $\displaystyle \frac{4}{\pi }\int_{0}^{\pi /2}{\frac{2+\sin ^{2}x}{\left( 2-\sin ^{2}x \right)^{2}}\,dx},$ which gives the desired result. $\displaystyle \blacksquare$

Hey K:

If you don't mind, could you please elaborate on your solution?. There is apparently something I am not up on and that is transforming series into integrals and the use of those combinatoric identities into integrals. Like you done with $\displaystyle \frac{1}{2^{3n}}\binom{2n}n=\frac{2}{\pi}\int{sin^ {2n}(x)}dx$. And how you changed that into that last integral you have. I would like to see how you came up with what you done. If it is too arduous, then forget about it. I don't want to put you out. Also, can you(or anyone else) recommend a book to learn more about this?.
I can come up with the series or something equivalent. I am not up on how you change that into a workable integral.
Not taking away from PH. I like his solution and I follow it.
They are both clever ways to go about it.
• Jun 21st 2008, 10:08 AM
Krizalid
Quote:

Originally Posted by galactus

I think there's no book dedicated to this, but notes about stuff like this do exist, dunno where exactly.

As for the problem:

$\displaystyle \sum\limits_{n\,=\,0}^{\infty }{\frac{(2n+1)!}{2^{3n}(n!)^{2}}}=\sum\limits_{n\, =\,0}^{\infty }{\frac{(2n+1)}{2^{n}}\cdot \frac{1}{2^{2n}}\binom{2n}n},$ hence the serie equals (assuming both of them do exist) $\displaystyle 2\sum\limits_{n\,=\,0}^{\infty }{\frac{n}{2^{n}}\cdot \frac{1}{2^{2n}}\binom{2n}n}+\sum\limits_{n\,=\,0} ^{\infty }{\frac{1}{2^{n}}\cdot \frac{1}{2^{2n}}\binom{2n}n}.$

Now insert the integral parameter and by simple geometric series application we have that the above equals

$\displaystyle \frac{4}{\pi }\int_{0}^{\pi /2}{\frac{\dfrac{\sin ^{2}x}{2}}{\left( 1-\dfrac{\sin ^{2}x}{2} \right)^{2}}\,dx}+\frac{4}{\pi }\int_{0}^{\pi /2}{\frac{dx}{2-\sin ^{2}x}},$

and the remaining algebra show us the simplified integral mentioned before.
• Jun 21st 2008, 10:15 AM
galactus
Thank you very much. I am going to research this more. Any sites or boks would be appreciated. You're a good and sharp egg.(Nod)
• Jun 21st 2008, 05:39 PM
ThePerfectHacker
Galactus I do not think there are any books on it. I tried to find them. But nobody ever wrote one. I dream that MHF would release an entire PDF on integration. That would MHF looks professional. But that dream is still far away.
• Jun 21st 2008, 05:44 PM
Krizalid
Quote:

Originally Posted by ThePerfectHacker

I dream that MHF would release an entire PDF on integration. That would MHF looks professional. But that dream is still far away.

• Jun 21st 2008, 05:48 PM
ThePerfectHacker
Quote:

Originally Posted by Krizalid

It is more than that. It is also a lot of time and work.
• Jun 22nd 2008, 12:20 AM
Isomorphism
Quote:

Originally Posted by ThePerfectHacker
Galactus I do not think there are any books on it. I tried to find them. But nobody ever wrote one. I dream that MHF would release an entire PDF on integration. That would MHF looks professional. But that dream is still far away.

Quote:

Originally Posted by Krizalid

This nice site lists many relations that have been popping up in the forum lately. We could take each one of the integrals, and either search the forum for that question, solve it ourselves or post it as a question. Then we can put it up on MHF wiki or make a pdf. That site was an example... we can get many such sites too.

Also mathwizard, jamesbond, perash and many others post good problems... We can make a pdf of it too.
• Jun 22nd 2008, 06:25 AM
galactus
Yes, that would be nice. A tutorial on trickier integration would be cool. With step-by-step procedures . I am decent enough at regular ol'

substitution, parts, etc when it comes to integration. But, some of the

clever ways Kriz, PH,

The

way Kriz uses a double integral and then switches the limits of integration is

cool; the various use of series and combinatorial identities to form and

integral, etc. It's ashame some of these techniques

are not taught in a calc class. I can do a lot of them, but Kriz and others

are certainly much better than me. You all have opened my eyes to

horizons. That is how one learns..by asking those who know. One reason I

like this site is because I can learn things from the learned. I am not so

arrogant that I am afraid to ask questions when I do not get something.

With math, there is always something new to learn. In contrast, I hope I

have learned some a few things.
• Jun 22nd 2008, 07:11 AM
ThePerfectHacker
Quote:

Originally Posted by galactus
It's ashame some of these techniques
are not taught in a calc class.

Most of the class is already lost in calculus. Start doing this sort of stuff and the class would be dead.
• Jun 22nd 2008, 07:41 AM
galactus
Yes, I reckon you are correct about that.