series sum?

Forum for the GRE subject test in mathematics.
hadimotamedi
Posts: 34
Joined: Thu Dec 30, 2010 4:36 am

series sum?

Postby hadimotamedi » Sat Jul 07, 2012 4:01 am

Dear All
Can you please let me know how can I calculate the summation of the following series when n approaches infinity?
Sn=1/2 + 1/3 + 2/9 + ... + an
where an = (2^(n-2))/(3^(n-1))
Thank you

Ell
Posts: 15
Joined: Fri Jan 20, 2012 3:46 am

Re: series sum?

Postby Ell » Sat Jul 07, 2012 3:43 pm

Starting from the second term, this is a geometric series with ratio = 2/3. Hence the summation equals the first term divided by 1 - the ratio.
So 1/3 + 2/9 + ... = (1/3)/(1-(2/3)) = 1
The entire series 1/2 + 1/3 + 2/9 + ... = 1/2 + 1 = 3/2

hadimotamedi
Posts: 34
Joined: Thu Dec 30, 2010 4:36 am

Re: series sum?

Postby hadimotamedi » Sat Jul 07, 2012 11:45 pm

Thank you very much for your help. I got the point.

vonLipwig
Posts: 51
Joined: Sat Mar 17, 2012 9:58 am

Re: series sum?

Postby vonLipwig » Sun Jul 08, 2012 5:21 am

You don't really need to treat the first term separately if you don't want to - the whole thing is a geometric series. That is, rather than multiplying 1/3 by 3 and adding 1/2, giving 3/2, you could just multiply the first term, 1/2, by 3, to get 3/2.

Ell
Posts: 15
Joined: Fri Jan 20, 2012 3:46 am

Re: series sum?

Postby Ell » Sun Jul 08, 2012 1:03 pm

Yes, you are right, the whole series is geometric. With the first term equal to 1/2, the whole sum is (1/2)/(1/3) = 3/2.




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