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Math Help - A Limit

  1. #1
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    A Limit

    Let I_n(p)=\frac{1}{n^{p+1}}\sum_{j=1}^n j^p, \ \ p \in \mathbb{R}. We know that if p > -1, then \lim_{n\to\infty}I_n(p)=\int_0^1 x^p \ dx = \frac{1}{p+1}.

    Question 1: Suppose p \in \mathbb{N}. Evaluate \lim_{n\to\infty} n \left(I_n(p) - \frac{1}{p+1} \right).

    Question 2: (maybe harder!) What is the above limit for -1 < p \in \mathbb{R} ?
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  2. #2
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    For Q.1 the answer 1/2 comes straight out of Faulhaber's formula. But that is independent of p, which makes it very tempting to think that the answer to Q.2 might be the same.
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    Quote Originally Posted by Opalg View Post

    ... which makes it very tempting to think that the answer to Q.2 might be the same.
    but (in Q.2) at p = 0 the limit is 0 not 1/2.
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  4. #4
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    Quote Originally Posted by NonCommAlg View Post
    but (in Q.2) at p = 0 the limit is 0 not 1/2.
    Hmmm yes, that's right. And if p<0 then the Riemann sum underestimates the integral, so the limit is going to be negative.

    I'll stop making conjectures at that point.
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