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Find $$\lim_{n\to\infty}\frac{1^p+3^p+\ldots+(2n+1)^p}{n^{p+1}}$$

I found a solution here which goes like this:

By Riemann sums, for any $p>-1$: $$ \frac{1}{n}\sum_{k=0}^{n}\left(\frac{2k+1}{n}\right)^p \xrightarrow{n\to +\infty}\int_{0}^{1}(2x)^p\,dx = \color{red}{\frac{2^p}{p+1}}.$$

I don't understand how you can change $\frac{2k+1}{n}$ to $2x$ there, like if there was $\frac{2k}{n}$ instead, I know we can easily change it to $2x$, but there is an extra $+1$, how can we igonre that term?

math_learner
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    use $2(k+1)\geq 2k+1 \geq 2k$ – psl2Z Apr 27 '24 at 16:32
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    At a more theoretical level you don't even need to justify it with inequalities. Here, using $2k+1$ in place of $2k$ is simply asserting that you are using a midpoint Riemann sum instead of a left or right handed sum. We often forget the more exotic varieties of Riemann sum when all of their limits do converge to the integral. – Ninad Munshi Apr 27 '24 at 17:03

3 Answers3

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$$\frac1n\sum_{k=0}^{n-1}\left(\frac{2k+1}n\right)^p\xrightarrow{n\to +\infty}\int_0^1(2x)^p\,dx$$ because it is a Riemann sum for $f(x)=(2x)^p$ and $x_k=\frac{k+1/2}n$ (so $x_{k+1}-x_k=\frac1n$). And the omitted term $\frac1n\left(\frac{2n+1}n\right)^p\sim\frac1n2^p$ tends to $0$.

Anne Bauval
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} & \color{#44f}{\lim_{n \to \infty} {1^{p} + 3^{p} + \cdots + \pars{2n + 1}^{p} \over n^{p+1}}}\ =\ \overbrace{\lim_{n \to \infty}\ {\pars{2n + 3}^{p} \over \pars{n + 1}^{\, p + 1} - n^{p + 1}}} ^{\substack{\ds{Stolz\mbox{-}Ces{\grave a}ro}\\[1mm] \ds{Theorem}\\[0.1mm]\mbox{}}} \\[5mm] = & \ 2^{p}\lim_{n \to \infty}\braces{{1 \over n^{p + 1}}\,{\,n^{p}\,\bracks{1 + 3/\pars{2n}}^{\, p} \over \pars{1 + 1/n}^{p + 1} - 1}} = 2^{p}\lim_{n \to \infty}\,\bracks{{1 \over n}\,{1 \over \pars{p + 1}/n}} \\[5mm] = & \ \bbx{\color{#44f}{2^{p} \over p + 1}} \\ & \end{align}

Felix Marin
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  • @AnneBauval I didn't remember that. It was 7 years ago. I'll delete this one tomorrow such that you can read this comment. – Felix Marin Apr 28 '24 at 04:49
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Not with Riemann sums but to go beyond the limit.

Let the numerator be $$\sum _{i=0}^n (2 i+1)^p$$Since we know that

$$\int(2i+1)^p\, di=\frac{(2 i+1)^{p+1}}{2( p+1)}$$ we can try the simples form of Euler-MacLaurin summation. It gives for

$$S_n=\sum _{i=0}^n (2 i+1)^p $$

$$S_n=\frac{(p-8) (p-6) (p-4) (p-2) p \left(p^3+4 p+10\right)}{9450 (p+1)}+$$ $$(2n+1)^p\,\Bigg(\frac{2 n+1}{2( p+1)}+\frac{1}{2}+\frac{p}{6 (2 n+1)}-\frac{(p-2) (p-1) p}{90 (2 n+1)^3}+O\left(\frac{1}{n^5}\right)\Bigg)\tag 1$$

that is to say, at the very first order $$\frac{S_n}{n^{p+1}} \sim \frac 1{2(p+1)} \left(\frac{2 n+1}{n}\right)^{p+1} \sim \frac {2^p}{p+1}$$ as already given by other answerers.

But $(1)$ is much more accurate for non infinite values of $n$ (as I did show it in the linked post).

Remark

It is interesting to notice that writing $$(2 i+1)^p=e^{p\log(2i+1) }$$ $$\sum _{i=0}^n e^{p\log(2i+1) }=2^p \left(\zeta \left(-p,\frac{3}{2}\right)-\zeta\left(-p,n+\frac{3}{2}\right)\right)$$ where appears the Hurwitz zeta function.

It could interesting to work this problem using the know asymptotics of this function.

Claude Leibovici
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