Infinite Product

Date: 12/01/97 at 02:13:53
From: Thang Nguyen
Subject: Infinite product

Hi, my name is Thang and I am having a hard time with this problem and 
wondering if you can help me. My problem is how to find the infinite 
product of this:

   3^1/3 * 9^1/9 * 27^1/27 * ..... (3^n)^1/3^n

I have tried to solve for a general formula like the geometric series 
where there is a general formula to work with, but that is where my 
problem started. Please help!

Thank you.

Date: 12/01/97 at 14:57:17
From: Doctor Bruce
Subject: Re: Infinite product

Hello Thang,

Try writing every term in your product as a power of 3.  So,

     3^(1/3)  =  3^(1/3)
     9^(1/9)  =  3^(2/9)
    27^(1/27) =  3^(3/27),

and so forth. Then, your infinite product becomes

     3^(1/3) * 3^(2/9) * 3^(3/27) * ...,

which can now be written as

     3^[1/3 + 2/9 + 3/27 + ... ]

where the exponent is an infinite sum. Now the task becomes how to
evaluate this infinite sum. This can be done by writing it as an 
infinite sum of infinite sums, like this

   1/3 + 1/9 + 1/27 + 1/81 + ...   =  1/2
         1/9 + 1/27 + 1/81 + ...   =  1/6
               1/27 + 1/81 + ...   =  1/18
                      1/81 + ...   =  1/54
                             ...   =  ...
__________________________________________

   1/3 + 2/9 + 3/27 + 4/81 + ...   =  (1/2)[1 + (1/3) + (1/9) + ...  ]

                                   =  3/4.

So it looks like the answer to your infinite product is  3^(3/4).

-Doctor Bruce,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   

Date: 12/11/97 at 16:02:51
From: Thang Nguyen
Subject: RE: calulus- infinite product

Hi Dr. Math,

I just want to say thank you so much for helping with this problem.

Thang Nguyen