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Math Help - Calculating series...

  1. #1
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    Calculating series...

    Hey, folks!

    How would you calculate the sum of (k^(j + n))/(m + k)^n, where summation variable is k and k = 1 to infinity?

    j, n, and m are positive integers.

    Thanks in advance.
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  2. #2
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    Solved. It doesn't converge.
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  3. #3
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    Mind showing your solution so this thread may be useful to others later on? Otherwise it will be deleted.
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  4. #4
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    Well, the necessary condition for series to converge is lim a_k -> 0 as k tends to infinity.

    In my case, a_k = (k^(j + n))/(m + k)^n = (k^j)*(1 - m/(m + k))^n.

    As k -> infinity, (1 - m/(m + k))^n -> 1, while k^j -> infinity.

    Which means a_k is unbounded. So, the necessary condition is not satisfied.
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  5. #5
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    Quote Originally Posted by curious View Post

    Well, the necessary condition for series to converge is lim a_k -> 0 as k tends to infinity.
    Consider \sum_{n\ge1}\frac1n, whereas \underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}=0, but the series diverges anyway.
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  6. #6
    Moo
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    Quote Originally Posted by Krizalid View Post
    Consider \sum_{n\ge1}\frac1n, whereas \underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}=0, but the series diverges anyway.
    He said necessary, not sufficient, so what he did is nothing else but correct.

    If \lim a_k \neq 0, then the series does not converge...
    This is the contrapositive of \sum a_k < + \infty \Rightarrow \lim a_k=0
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  7. #7
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    Yeah, sure. It's only the necessary condition.

    That is, if a_k -> 0, it doesn't mean that the series will converge.

    But if a_k -> "something other than 0", then definitely the series won't converge.
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