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

Question

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.

Answer 1

Here's an alternative method that Lord Shark the Unknown suggested (as Oscar Lanzi pointed out, this only works for integer $a,b$, though Claude Leibovici's answer is essentially this only generalized with asymptotics for any $a,b$):

We have by Faulhaber's formula that $$ \sum_{k=1}^nk^a=\frac{n^{a+1}}{a+1}+P_a(n) $$ where $P_a$ denotes some polynomial of degree $a$. So we get that \begin{align} \lim_{n\rightarrow\infty}n^{b-a}\frac{\sum_{k=1}^nk^a}{\sum_{r=1}^nr^b}=\lim_{n\rightarrow\infty}\frac{n^{-(a+1)}\left(\frac{n^{a+1}}{a+1}+P_a(n)\right)}{n^{-(b+1)}\left(\frac{n^{b+1}}{b+1}+P_b(n)\right)}=\lim_{n\rightarrow\infty}\frac{\frac{1} {a+1}+n^{-(a+1)}P_a(n)}{\frac{1}{b+1}+n^{-(b+1)}P_b(n)} \end{align} The expressions $n^{-(c+1)}P_c(n)$ tend to $0$ as $n\rightarrow\infty$ since the degree of $P_c$ is $c$. Therefore: $$ \lim_{n\rightarrow\infty}\frac{\frac{1} {a+1}+n^{-(a+1)}P_a(n)}{\frac{1}{b+1}+n^{-(b+1)}P_b(n)}=\lim_{n\rightarrow\infty}\frac{\frac{1} {a+1}+0}{\frac{1}{b+1}+0}=\frac{b+1}{a+1} $$

Answer 2

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).

Answer 3

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.