For $a,b>-1$
$$\lim_{n\to \infty} n^{b-a}\frac{1^a+2^a+\cdots +n^a}{1^b+2^b+\cdots +n^b}$$
I am really confused in this one. I tried to calculate it but the answer comes out to be $1$ as I divided the numerator and denominator with $n^a$ and $n^b$ respectively. But the answer is wrong.
Please help.
Best Answer
I assume this is the question
$$\displaystyle \lim_{n\to \infty} n^{b-a}\dfrac{1^a+2^a+....+n^a}{1^b+2^b+....+n^b}$$
This can be written as
$$\displaystyle \lim_{n\to \infty} \dfrac{n^b}{n^a} \dfrac{\sum_{r=1}^n r^a}{\sum_{k=1}^n k^b}$$
$$\displaystyle \lim_{n\to \infty} \dfrac{\sum_{r=1}^n \left(\dfrac{r}{n}\right)^a}{\sum_{k=1}^n \left(\dfrac{k}{n}\right)^b}$$
Or
$$\displaystyle \lim_{n\to \infty} \dfrac{\frac{1}{n}\sum_{r=1}^n \left(\dfrac{r}{n}\right)^a}{\frac{1}{n}\sum_{k=1}^n \left(\dfrac{k}{n}\right)^b}$$
$$\displaystyle \dfrac{\int_0 ^1 x^a dx}{\int_0 ^1 x^b dx}$$
$$\displaystyle =\dfrac{b+1}{a+1}$$
where the conditions $a>-1, b>-1$ guarantees that the integrals converge (without this convergence the calculation with Riemann sums does not work).