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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.

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    $\begingroup$ Use $\sum_{k=1}^n k^a\sim(a+1)^{-1}n^{a+1}$. $\endgroup$ – Angina Seng Jul 13 '19 at 11:27
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    $\begingroup$ I am sorry but I did not understand that. Can you please repeat? $\endgroup$ – Asad Ahmad Jul 13 '19 at 11:28
  • $\begingroup$ Cesaro-Stolz is the key here. Write the expression as $f(a, n) /f(b, n) $ where $f(x, n) =n^{-x-1}\sum_{k=1}^{n}k^x$. $\endgroup$ – Paramanand Singh Jul 14 '19 at 4:00
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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: $$ \lim_{n\rightarrow\infty}\frac{\frac{1} {a+1}+n^{-(a+1)}P_a(n)}{\frac{1}{b+1}+n^{-(b+1)}P_b(n)}=\lim_{n\rightarrow\infty}\frac{\frac{1} {a+1}+0}{\frac{1}{b+1}+0}=\frac{b+1}{a+1} $$

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    $\begingroup$ The limitation here is that when the "remainder" in the Faulhaber formula is called polynomial, $a$ and $b$ are assumed whole numbers. But the result also applies for non-whole number values of the parameters (in the proper range), so we need a broader method to capture the full result. $\endgroup$ – Oscar Lanzi Jul 13 '19 at 15:34
  • $\begingroup$ Right, I missed that possibly $a,b\in\mathbb{R}-\mathbb{Z}$. Have edited. $\endgroup$ – J_P Jul 13 '19 at 15:40
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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).

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  • $\begingroup$ Thanks a lot! Just asking if there is any other way of solving this question without the use of sum/integral? $\endgroup$ – Asad Ahmad Jul 13 '19 at 11:40
  • $\begingroup$ I am not sure about any other method $\endgroup$ – Akash Karnatak Jul 13 '19 at 11:43
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    $\begingroup$ nice intuition that of transforming into Riemann sum ! $\endgroup$ – G Cab Jul 13 '19 at 12:07
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    $\begingroup$ Added a note about the integrals converging. The limit as rendered here does not work otherwise. In the problem statement convergence is forced by $a,b>-1$. $\endgroup$ – Oscar Lanzi Jul 13 '19 at 15:30
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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.

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  • $\begingroup$ Nice solution! Though didn't you miss the $n^{-a}\zeta(-a)$ term in the last line? If $a<0$ or $b<0$ that can change the asymptotic formula at the end. $\endgroup$ – J_P Jul 13 '19 at 15:51
  • $\begingroup$ @J_P. Good point ! $\endgroup$ – Claude Leibovici Jul 14 '19 at 2:53

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