# A Limit

• Jan 10th 2009, 01:24 AM
NonCommAlg
A Limit
Let $\displaystyle I_n(p)=\frac{1}{n^{p+1}}\sum_{j=1}^n j^p, \ \ p \in \mathbb{R}.$ We know that if $\displaystyle p > -1,$ then $\displaystyle \lim_{n\to\infty}I_n(p)=\int_0^1 x^p \ dx = \frac{1}{p+1}.$

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

Question 2: (maybe harder!) What is the above limit for $\displaystyle -1 < p \in \mathbb{R}$ ?
• Jan 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)
• Jan 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.
• Jan 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)