Finding $\lim_{n\to \infty} n^{b-a}\frac{1^a+2^a+\cdots +n^a}{1^b+2^b+\cdots +n^b}$ 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.

Another similar solution using generalized harmonic numbers
$$S_a=\sum_{i=1}^n i^a=H_n^{(-a)}$$ Using asymptotics
$$S_a=n^a \left(\frac{n}{a+1}+\frac{1}{2}+\frac{a}{12
   n}+O\left(\frac{1}{n^3}\right)\right)+\zeta (-a)$$
$$n^{-a} S_a=\frac{n}{a+1}+\frac{1}{2}+\frac{a}{12
   n}+O\left(\frac{1}{n^3}\right)+n^{-a}\zeta (-a)$$
$$n^{b-a}\frac{S_a}{S_b}=\frac{n^{-a}S_a}{n^{-b}S_b}\sim \frac{\frac{n}{a+1}+\frac{1}{2}+\frac{a}{12
   n} }{\frac{n}{b+1}+\frac{1}{2}+\frac{b}{12
   n} }=\frac{b+1}{a+1}+\frac{(b+1) (a-b)}{2 (a+1)
   n}+O\left(\frac{1}{n^2}\right)$$ which shows the limit and also how it is approached.
