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 .[/color] 
May 16th, 2012, 05:26 AM  #2 
Senior Member Joined: Oct 2011 From: Belgium Posts: 522 Thanks: 0  Re: Interesting Series. 
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:
and next?  
May 16th, 2012, 12:47 PM  #6  
Global Moderator Joined: May 2007 Posts: 6,787 Thanks: 708  Re: Interesting Series. Quote:
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:
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 . Let us sum until a big N and then let's make N tends to infinite. 1)First we realize that : Now you can multiply by the other odd terms and you get a part that is equal to 2) Then we see that 3) So we can write that 4) Now using the Stirling formula we have that as N tends to infinite. 5) So the limit is 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 no??


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