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Let $p$ be a real number. Evaluate $\displaystyle \lim_{n \to \infty} \sum_{k=1}^n \dfrac{k^p}{n^{p+1}}$.

I think this depends on the value of $p$ because then large $n$ would mean small $n^{p+1}$ for small $p$ etc. Should I break this up into cases where $p > 1$ and $p \leq 1$? In any case, since the sum converges we have $$\lim_{n \to \infty} \sum_{k=1}^n \dfrac{k^p}{n^{p+1}} = \lim_{n \to \infty}\dfrac{1}{n^{p+1}} \sum_{k=1}^n k^p.$$ I am not sure how to continue since expanding $\displaystyle \sum_{k=1}^n k^p$ may be hard.

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  • $\begingroup$ It looks like integrating $x^p$ from $0$ to $1$ to me. Hope this might help $\endgroup$
    – k99731
    Commented Apr 26, 2016 at 3:25

1 Answer 1

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HINT:

What is the Riemann sum for $\int_0^1 x^{p}\,dx$?

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  • $\begingroup$ That is $\lim_{\Delta{x_i} \to 0}\sum_{k = 0}^{\infty}(x^*)^p(\Delta{x_i})$. How does that help? $\endgroup$
    – user19405892
    Commented Apr 26, 2016 at 3:35
  • $\begingroup$ Rewrite the limit as: $$\lim\limits_{n\to\infty}\sum\limits_{k=1}^n\frac{1}{n}\cdot \left( \frac{k}{n} \right )^p $$ Can you relate this to the formula for the Riemann sum you just posted? $\endgroup$
    – Fimpellizzeri
    Commented Apr 26, 2016 at 3:41
  • $\begingroup$ @Fimpellizieri I see the width length $\Delta_{x_i} = \dfrac{1}{n}$, but I don't see how to relate the other term. $\endgroup$
    – user19405892
    Commented Apr 26, 2016 at 3:52
  • $\begingroup$ @user19405892 The Riemann sum for $\int_0^1 x^p\,dx$ is $\lim_{n\to \infty}\frac1n \sum_{k=1}^n (k/n)^p$. $\endgroup$
    – Mark Viola
    Commented Apr 26, 2016 at 3:56
  • $\begingroup$ So the answer is $\frac{1}{2}$? $\endgroup$
    – user19405892
    Commented Apr 26, 2016 at 4:04

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