Infinite ProductDate: 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 |
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