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Math Help - Convergence- proving bounded

  1. #1
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    Convergence- proving bounded

    Given sequence, a= 1/1^2+1/2^2+1/3^2+.......+1/n^2

    (1) I have to prove for natural numbers n, such that n is greater than or equal to 2,

    a is less than or equal to 1+1/1*2+1/2*3+1/3*4+.....+1/(n-1)*n

    (2) Prove that a is bounded for natural numbers

    I've been given a hint in the question to use the fact that for n greater than or equal to 2, 1/(n-1)n=1/(n-1) -1/n

    Anyone have any ideas? Anything would be very much appreciated
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  2. #2
    Senior Member roninpro's Avatar
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    Hello.

    In part one, you can just compare terms directly. The n-term in your first series is \frac{1}{n^2}, and the n-term in your second series is \frac{1}{(n-1)n}. Is it true that \frac{1}{n^2}\leq\frac{1}{(n-1)n} (for n>2)? If so, then you have 1+\frac{1}{2^2}+\ldots+\frac{1}{n^2}<1+\frac{1}{1\  cdot 2}+\ldots+\frac{1}{(n-1)n}, and therefore, the first series is bounded by the second.

    In part two, we can try to sum up the second series to put a bound on the first. Try writing it down using your hint:

    1+\frac{1}{1\cdot  2}+\ldots+\frac{1}{(n-1)n}=1+\frac{1}{2}+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+\ldots+(\frac{1}{n-2}-\frac{1}{n-1})+(\frac{1}{n-1}-\frac{1}{n})

    Anything nice happen? What happens when you take n\to \infty? Use this to conclude.
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