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What is the result of the next limit:

$$ \lim_{n \to \infty} \frac{ \sum^n_{i=1} i^k}{n^{k+1}},\ k \in \mathbb{R} $$

Why (theorem)?

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You can apply Lemma Stolz-Cesaro: $$ \lim_{n \to \infty} \frac{ \sum^n_{i=1} i^k}{n^{k+1}}= \lim_{n \to \infty} \frac{ (n+1)^k}{(n+1)^{k+1}-n^{k+1}}=\frac{1}{k+1}$$

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  • $\begingroup$ How do you get $ 1/(k+1) $? $\endgroup$ – David Aug 28 '13 at 21:20
  • $\begingroup$ @David as n gets huge the fraction in the limit basically becomes ... logic doesn't apply, Look closely. $(n+1)^{k+1}$ and $n^{k+1}$ are not actually that far apart. Yeah they are for k=1 and n from 4 to 5 say (25-16 = 9, not a small amount!) but after a while they come close to each other (look at the graph of their ratio for some powers, it is interesting) but they are not the same, the denominator tends towards 0 from above (as k gets large) but the numerator also gets huge. So the thing should get huge! If you do 1/ limit you get something that looks like 1 for huge k. $\endgroup$ – Alec Teal Apr 24 '14 at 7:34
  • $\begingroup$ @David that contradicts (it looks like $n+1-n\frac{n^k}{(n+1)^k}$ and that last fraction tends towards 1 from below. So we have tends towards 0 from above +1, which is 1, 1/1 is 1. This is why you need to be careful with things that go near 0 in the denominator. 1 and $+\infty$ are both wrong for huge k! So that $\frac{1}{k+1}$ comes from some special place, it isn't obvious. $\endgroup$ – Alec Teal Apr 24 '14 at 7:38
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This is a Riemann sum, which in this limit takes on the value of an integral:

$$\lim_{n\to\infty} \frac{1}{n} \sum_{i=1}^n \frac{i^k}{n^k} = \int_0^1 dx \, x^k = \frac{1}{k+1}$$

so long as $k \gt -1$.

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    $\begingroup$ As long as $k > -1$. $\endgroup$ – D. Thomine Aug 28 '13 at 18:01
  • $\begingroup$ Because the LHS goes to $+ \infty$, while the RHS is negative? Sure, you can probably define your answer as some kind of sum of a divergent serie, but that's not the answer to David's problem. $\endgroup$ – D. Thomine Aug 28 '13 at 18:05
  • $\begingroup$ Could one apply the Stolz–Cesàro theorem? $\endgroup$ – David Aug 28 '13 at 18:06
  • $\begingroup$ The numerator converges to a positive constant. The denominator goes to $0$. The answer is straightforward. $\endgroup$ – D. Thomine Aug 28 '13 at 18:07
  • $\begingroup$ @D.Thomine: what was I thinking? Yes, of course, thank you. $\endgroup$ – Ron Gordon Aug 28 '13 at 18:07
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Just to clarify Ron's answer, the result you're looking for is the following: $$\lim_{n \to \infty} \frac{K}{n}\sum_{i=1}^n f\left(\frac{Ki}{n}\right)=\int_0^K f(x) \, \mathrm{d}x. $$ In your case, $f(x) = x^k$, and $K=1$. The trick here is that whenever you deal with the limit of a sum, factor outside the sum $1/n$ and see whether you can group the sum term to reduce to the Riemann sum expression. If you can, switch to integral form.

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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\lim_{n\ \to\ \infty}{\sum^{n}_{i\ =\ 1}i^{k} \over n^{k + 1}}:\ {\large ?}. \qquad k \in {\mathbb R}}$.


  1. $\ds{\large\dsc{k < -1}:}$ \begin{align}&\lim_{n\ \to\ \infty}{\sum^{n}_{i\ =\ 1}i^{k} \over n^{k + 1}} =\lim_{n\ \to\ \infty}\pars{% n^{\verts{k + 1}}\sum^{n}_{i\ =\ 1}{1 \over i^{\verts{k}}}} =\color{#66f}{\large\infty} \end{align} Note that $\ds{\lim_{n\ \to\ \infty}\sum^{n}_{i\ =\ 1}{1 \over i^{\verts{k}}}=\zeta\pars{\verts{k}}}$
  2. $\ds{\large \dsc{k = -1}:}$ \begin{align}&\lim_{n\ \to\ \infty}{\sum^{n}_{i\ =\ 1}i^{k} \over n^{k + 1}} =\lim_{n\ \to\ \infty}\sum^{n}_{i\ =\ 1}{1 \over i}=\color{#66f}{\large\infty} \end{align} Note that $\ds{\sum^{n}_{i\ =\ 1}{1 \over i}=H_{n}}$ where $\ds{H_{n}}$ is the Harmonic Number which diverges logarithmically $\ds{\pars{~H_{n} \sim \ln\pars{n}\ \mbox{when}\ n \gg 1~}}$.
  3. $\ds{\large \dsc{k > -1}:}$ \begin{align}\lim_{n\ \to\ \infty}{\sum^{n}_{i\ =\ 1}i^{k} \over n^{k + 1}} &=\lim_{n\ \to\ \infty} {\sum^{n + 1}_{i\ =\ 1}i^{k} - \sum^{n}_{i\ =\ 1}i^{k}\over \pars{n + 1}^{k + 1} - n^{k + 1}} =\lim_{n\ \to\ \infty} {\pars{n + 1}^{k} \over \pars{n + 1}^{k + 1} - n^{k + 1}} =\color{#66f}{\large{1 \over k + 1}} \end{align}
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