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Thread: 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 $\displaystyle \frac{3^n + 2^n}{6^n} = \frac{3^n + 2^n}{3^n2^n}= \frac{1}{3^n} + \frac{1}{2^n}$

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

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

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


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

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


    The first series has sum

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

    The second series has sum

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


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

    You could also do that by noting that $\displaystyle \sum_{n=1}^{\infty}r^n$ 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:
    $\displaystyle \frac{1}{1- \frac{1}{3}}- 1= \frac{3}{2}- 1= \frac{1}{2}$ and $\displaystyle \frac{1}{1- \frac{1}{2}}- 1= 2- 1= 1$.

    serenalee, if you got $\displaystyle \frac{3}{2}+ 2= \frac{7}{2}$ instead of $\displaystyle \frac{1}{2}+ 1= \frac{3}{2}$, 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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