A Limit

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January 10th 2009, 01:24 AM
NonCommAlg

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 $-1 < p \in \mathbb{R}$ ?
January 10th 2009, 05:05 AM
Opalg

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. (Wondering)
January 10th 2009, 09:02 AM
NonCommAlg

Quote:

Originally Posted by Opalg

... which makes it very tempting to think that the answer to Q.2 might be the same. (Wondering)

but (in Q.2) at p = 0 the limit is 0 not 1/2.
January 10th 2009, 10:50 AM
Opalg

Quote:

Originally Posted by NonCommAlg

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. (Lipssealed)

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