My Math Forum Interesting Series.

 Real Analysis Real Analysis Math Forum

 May 16th, 2012, 04:06 AM #1 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Interesting Series. [color=#000000]Compute $\sum_{n=0}^{\infty}\left(1+n\log\left(\frac{2n}{2n +2}\right)+\log\left(\frac{2n+1}{2n+2}\right)\righ t)$ .[/color]
 May 16th, 2012, 05:26 AM #2 Senior Member   Joined: Oct 2011 From: Belgium Posts: 522 Thanks: 0 Re: Interesting Series. $=\lim_{k \to \infty} \sum_{n=0}^{k}\left(1+n\log\left(\frac{2n}{2n+2}\r ight)+\log\left(\frac{2n+1}{2n+2}\right)\right)$ $=\lim_{k \to \infty} \sum_{n=0}^{k}\left(1+n\log\left(2n\right)+\log\le ft(2n+1\right)-\left(n+1\right)\log\left(2n+2\right)\right)$ $=\lim_{k \to \infty} \sum_{n=0}^{k}\left(\log\left(2n+1\right)\right)+k +1-\left(k+1\right)\log\left(2k+2\right)$ $=???$
 May 16th, 2012, 06:00 AM #3 Senior Member   Joined: May 2011 Posts: 501 Thanks: 6 Re: Interesting Series. ....
 May 16th, 2012, 10:13 AM #4 Senior Member   Joined: Nov 2011 Posts: 595 Thanks: 16 Re: Interesting Series. Probably you should compute the exponential of the series and then the infinite product might be reducible...
May 16th, 2012, 10:57 AM   #5
Senior Member

Joined: Oct 2011
From: Belgium

Posts: 522
Thanks: 0

Re: Interesting Series.

Quote:
 Originally Posted by Dougy Probably you should compute the exponential of the series and then the infinite product might be reducible...
you mean

$\prod_{n=0}^{\infty}\left(e\cdot \left(\frac{2n}{2n+2}\right)^n \cdot \left(\frac{2n+1}{2n+2}\right)\right)$

and next?

May 16th, 2012, 12:47 PM   #6
Global Moderator

Joined: May 2007

Posts: 6,787
Thanks: 708

Re: Interesting Series.

Quote:
 Originally Posted by ZardoZ [color=#000000]Compute $\sum_{n=0}^{\infty}\left(1+n\log\left(\frac{2n}{2n +2}\right)+\log\left(\frac{2n+1}{2n+2}\right)\righ t)$ .[/color]
It looks like the series should telescope (cancellation between adjacent terms).
Add the kth term to the (k+1)th term and see what happens.

May 16th, 2012, 12:57 PM   #7
Senior Member

Joined: Oct 2011
From: Belgium

Posts: 522
Thanks: 0

Re: Interesting Series.

Quote:
Originally Posted by mathman
Quote:
 Originally Posted by ZardoZ [color=#000000]Compute $\sum_{n=0}^{\infty}\left(1+n\log\left(\frac{2n}{2n +2}\right)+\log\left(\frac{2n+1}{2n+2}\right)\righ t)$ .[/color]
It looks like the series should telescope (cancellation between adjacent terms).
Add the kth term to the (k+1)th term and see what happens.
$=1+n\log\left(\frac{2n}{2n+2}\right)+\log\left(\fr ac{2n+1}{2n+2}\right)+1+\left(n+1\right)\log\left( \frac{2n+2}{2n+4}\right)+\log\left(\frac{2n+3}{2n+ 4}\right)$

Not much that happens...

 May 16th, 2012, 01:55 PM #8 Senior Member   Joined: Nov 2011 Posts: 595 Thanks: 16 Re: Interesting Series. That's a real headache but I think I got it! So use the product that wnlv wrote $\prod_{n=0}^{\infty}\left(e\cdot \left(\frac{2n} {2n+2}\right)^n \cdot \left(\frac{2n+1}{2n+2}\ right)\right)$. Let us sum until a big N and then let's make N tends to infinite. 1)First we realize that : $\prod_{n=0}^{N}\cdot \left(\frac{2n}{2n+2}\right )^n=1\times 2\times 4\times ...\times \fra{2N}{(2N+2)^N}$ Now you can multiply by the other odd terms $\prod_{n=0}^{N}(2n+1)$ and you get a part that is equal to $\frac{(2N+1)!}{(2N+2)^N}$ 2) Then we see that $\prod_{n=0}^{N}\frac{1}{2n+2}=\frac{1}{2^{N+1}(N+1 )!}$ 3) So we can write that $\prod_{n=0}^{N}\left(e\cdot \left(\frac{2n}{2n+2 }\right)^n \cdot \left(\frac{2n+1}{2n+2}\right )\right)={(\frac{e}{2})}^{N+1}\frac{(2N+1)!}{(2N+2 )^N(N+1)!}$ 4) Now using the Stirling formula we have that $\frac{e^{N+1}(2N+1)!}{2^{N+1}(N+1)!(2N+2)^N}\to e\sqrt{2}$ as N tends to infinite. 5) So the limit is $\log(\sqrt 2)$ Edited
 May 16th, 2012, 02:08 PM #9 Senior Member   Joined: Oct 2011 From: Belgium Posts: 522 Thanks: 0 Re: Interesting Series. removed
 May 16th, 2012, 02:26 PM #10 Senior Member   Joined: Nov 2011 Posts: 595 Thanks: 16 Re: Interesting Series. are you sure? look for instance until N=5 you have $\frac{1 \times 2^1 \times 4^2 \times 6^3 \times 8^4 \times 10^5}{4^1 \times 6^2 \times 8^3 \times 10 ^4 \times 12 ^5}=2\times 4 \times 6 \times 8 \times 10 /12^5$ no??

 Tags interesting, series

,

### math interesting series

Click on a term to search for related topics.
 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post eddybob123 Math Events 2 October 9th, 2013 03:16 PM benten Algebra 10 May 14th, 2013 08:36 PM fride Algebra 1 February 23rd, 2013 09:49 PM galactus Real Analysis 2 May 24th, 2012 01:11 PM Eval New Users 6 April 22nd, 2011 05:54 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top