$\lim\limits_{r\rightarrow \infty , 0} \left(\frac{\sum_1^n i^r/ n}{\sum_1^{n+1} i^r/(n+1)}\right)^{1/r}$?

Edited

please help me! how can I evaluate: $$\lim_{r\rightarrow 0} \left(\frac{\sum_1^n i^r/ n}{\sum_1^{n+1} i^r/(n+1)}\right)^{1/r}$$ and

$$\lim_{r\rightarrow \infty} \left(\frac{\sum_1^n i^r/ n}{\sum_1^{n+1} i^r/(n+1)}\right)^{1/r}$$

-
What does $r \to \infty , 0$ mean? Are we evaluating two problems here? One when $r \to \infty$ and the other when $r \to 0$? – Gautam Shenoy Mar 11 '13 at 10:00
yes, it is two problems. one for $r\rightarrow \infty$ and the other for $r\rightarrow 0$ – Math 1988 Mar 11 '13 at 10:27
limits are evaluated not solved – Arjang Mar 11 '13 at 11:32

Hints:

Focus on the expression $\left(\frac{1}{n}\sum_{i=1}^n i^r\right)^{1/r}$.

For $r$ large, note $n^{r-1}\leq \frac{1}{n}\sum_{i=1}^n i^r\leq n^r$.

For $r$ small, try taking logs and applying l'Hospital's rule.

-

I'll solve the 0 case. $$\lim_{r\to 0}\left(\dfrac{(n+1)(\sum_1^ni^r)}{n\sum_1^{n+1}i^r}\right)^{\frac 1 r}=\lim_{r\to 0} \left(\dfrac{1 +r\dfrac{\log n!}{n}}{1 +r\dfrac{\log n+1!}{n+1}}\right)^{\frac 1r}=\exp\left({\dfrac{\log n!}{n}-\dfrac{\log (n+1)!}{n+1}}\right)$$ This is all by taking a $1^{st}$ order approximation.

-