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Math Help - Question about geometric progression

  1. #1
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    Question about geometric progression

    Given that sum from n=1 to infinity ( 3^n + 2^n) / 6^n.

    Was asked to determine if it's convergent or divergent. If convergent, find its sum.

    What i did was

    (3/6)^n + (2/6)^n
    = (1/2)^n + (1/3)^n

    since both 1/2 and 1/3 are less than 1. so it's convergent.
    hence I used a/1-r to find the sum to infinity for both 1/2 and 1/3 seperate cases and added up both the sum. But it was wrong. why?
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  2. #2
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    Quote Originally Posted by serenalee View Post
    Given that sum from n=1 to infinity ( 3^n + 2^n) / 6^n.

    Was asked to determine if it's convergent or divergent. If convergent, find its sum.

    What i did was

    (3/6)^n + (2/6)^n
    = (1/2)^n + (1/3)^n

    since both 1/2 and 1/3 are less than 1. so it's convergent.
    hence I used a/1-r to find the sum to infinity for both 1/2 and 1/3 seperate cases and added up both the sum. But it was wrong. why?
    Note that \frac{3^n + 2^n}{6^n} = \frac{3^n + 2^n}{3^n2^n}= \frac{1}{3^n} + \frac{1}{2^n}

    So \sum_{i = 1}^{\infty}\left(\frac{3^n + 2^n}{6^n}\right) = \sum_{i = 1}^{\infty}\left(\frac{1}{3^n} + \frac{1}{2^n}\right)

     = \sum_{i = 1}^{\infty}\frac{1}{3^n} + \sum_{i = 1}^{\infty}\frac{1}{2^n}

     = \sum_{i = 1}^{\infty}\left(\frac{1}{3}\right)^n + \sum_{i = 1}^{\infty}\left(\frac{1}{2}\right)^n.


    Each of these is a geometric series. The first has a = \frac{1}{3} and r = \frac{1}{3}. The second has a = \frac{1}{2} and r = \frac{1}{2}.

    Since each has |r| < 1, both series are convergent, so the whole series is convergent.


    The first series has sum

    S_\infty = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{1}{2}.

    The second series has sum

    S_\infty = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = 1.


    So \sum_{i = 1}^{\infty}\left(\frac{3^n + 2^n}{6^n}\right) = \frac{1}{2} + 1 = \frac{3}{2}.
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  3. #3
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    Critical point: notice that the sums, as shown by Prove It: \sum_{n=1}^\infty \frac{1}{3^n} and \sum_{n=1}^\infty \frac{1}{2^n}, start at n= 1. The formula \sum_{n=0}^\infty ar^n= \frac{a}{1- r} has the sum starting at n= 0. Prove It handled that by taking a= 1/3 and 1/2 so they were \frac{1}{3}\sum_{n=0}^\infty \frac{1}{3}^n and \frac{1}{2}\sum_{n=0}^\infty \frac{1}{2^n}.

    You could also do that by noting that \sum_{n=1}^{\infty}r^n is just missing the "n= 0" term which is 1. So just use the standard formula, with a= 1, r= 1/3 or 1/2, and subtract 1:
    \frac{1}{1- \frac{1}{3}}- 1= \frac{3}{2}- 1= \frac{1}{2} and \frac{1}{1- \frac{1}{2}}- 1= 2- 1= 1.

    serenalee, if you got \frac{3}{2}+ 2= \frac{7}{2} instead of \frac{1}{2}+ 1= \frac{3}{2}, that was your error.

    (In the future, if you are asking why your answer was wrong, it would be a good idea to tell us what your answer was!)
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