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Math Help - Determine whether the following series converges or diverges.

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    Determine whether the following series converges or diverges.

    Determine whether the following series converges or diverges


    ∑ [(-3/∏)^n + (e/9)^(n+1)]
    n=1

    If it converges, find its sum.


    I don't know how to go about solving this one, can someone explain the steps that need to be taken?

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    Re: Determine whether the following series converges or diverges.

    Distribute the sum over both terms, and if the series converges, both of those terms should converge.

    If ∑an and ∑bn are convergent series, then so are the series ∑can (where c is a constant)

    Then ∑(an + bn) =
    an + ∑bn
    Last edited by sjmiller; September 30th 2012 at 12:40 PM.
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    Re: Determine whether the following series converges or diverges.

    Quote Originally Posted by sjmiller View Post
    Distribute the sum over both terms, and if the series converges, both of those terms should converge.

    If ∑an and ∑bn are convergent series, then so are the series ∑can (where c is a constant)

    Then ∑(an + bn) =
    an + ∑bn
    Im still not sure where to go from there. This is a geometric series, correct?
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    Re: Determine whether the following series converges or diverges.

    Quote Originally Posted by sjmiller View Post
    Distribute the sum over both terms, and if the series converges, both of those terms should converge.

    If ∑an and ∑bn are convergent series, then so are the series ∑can (where c is a constant)

    Then ∑(an + bn) =
    an + ∑bn
    Quote Originally Posted by Preston019 View Post
    Im still not sure where to go from there. This is a geometric series, correct?
    Correct they are both geometric series. Remember the sum of such a series ∑ arn-1 = a/(1-r) provided that |r|<1
    You need to get them into a form where you can easily determine the values of a and r. If |r| > 1 then the series diverges.
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    Re: Determine whether the following series converges or diverges.

    Quote Originally Posted by sjmiller View Post
    Correct they are both geometric series. Remember the sum of such a series ∑ arn-1 = a/(1-r) provided that |r|<1
    You need to get them into a form where you can easily determine the values of a and r. If |r| > 1 then the series diverges.
    Okay, so is this true: ∑ ar^(-n-1) = a/(1-r)^-1

    And how would I get ∑(-3/∏)^n into the proper form?
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    Re: Determine whether the following series converges or diverges.

    Quote Originally Posted by Preston019 View Post
    Okay, so is this true: ∑ ar^(-n-1) = a/(1-r)^-1

    And how would I get ∑(-3/∏)^n into the proper form?
    No that does not follow from the original formula.

    Note that (-3/pi)^n can be written as (-3/pi)(-3/pi)^(n-1) where a = (-3/pi) and r = (-3/pi) and |r| < 1 so it indeed converges.
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    Re: Determine whether the following series converges or diverges.

    Quote Originally Posted by sjmiller View Post
    No that does not follow from the original formula.

    Note that (-3/pi)^n can be written as (-3/pi)(-3/pi)^(n-1) where a = (-3/pi) and r = (-3/pi) and |r| < 1 so it indeed converges.
    So, then how do I put ∑(e/9)^(n+1) into the proper form?
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    Re: Determine whether the following series converges or diverges.

    Quote Originally Posted by Preston019 View Post
    Determine whether the following series converges or diverges

    ∑ [(-3/∏)^n + (e/9)^(n+1)]
    n=1
    If it converges, find its sum.
    Lets do the using partial sums.
    If S_n  = \sum\limits_{k = 1}^n {\left[ {\left( {\frac{{ - 3}}{\pi }} \right)^k  + \left( {\frac{e}{9}} \right)^k } \right]} would you agree that S_n=U_n+V_n where U_n  = \sum\limits_{k = 1}^n {\left[ {\left( {\frac{{ - 3}}{\pi }} \right)^k } \right]} and  V_n  = \sum\limits_{k = 1}^n { {\left( {\frac{e}{9}} \right)^k } }

    Does each of U_n~\&~V_n converge? If so, the sum of their limits is the sum of the original series.
    Last edited by Plato; September 30th 2012 at 03:28 PM.
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    Re: Determine whether the following series converges or diverges.

    Quote Originally Posted by Preston019 View Post
    So, then how do I put ∑(e/9)^(n+1) into the proper form?
    Same idea as the prior, instead of decreasing n by 1 we now want to decrease n by 2.

    (e/9)^(n-1) = (e/9)^2(e/9)^(n-1)

    Not that if you add the exponents together you get 2 + (n - 1) = n + 1
    This is always your goal with a geometric series. Always check that the absolute value of r is less than 1, because if r is greater than 1 the series diverges.
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