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 $\displaystyle -1 < p \in \mathbb{R}$ ?