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

Question

I'm trying to solve the following problem:

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

What I've got so far:

My idea is to use Stolz-Cesaro theorem, which implies that:

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

From there:

$$\frac{(2n+3)^p}{(n+1)^{p+1}-n^{p+1}}=\frac{(2n+3)^p}{(2n+1)^{\frac{p+1}{2}}}$$

So, if my calculations are correct I only have to find the limit of the term in the RHS on the last line, that's unless the approach is somewhere entirely different.

Thanks in advance!

Answer 1

You are on the right track: as $n\to +\infty$, $$\frac{(2n+3)^p}{(n+1)^{p+1}-n^{p+1}}=\frac{2^pn^p(1+\frac{3}{2n})^p}{n^{p+1}\left(\left(1+\frac{1}{n}\right)^{p+1}-1\right)}=\frac{2^p(1+o(1))}{n\left(1+\frac{p+1}{n}+o(1/n)-1\right)}\to {\frac{2^p}{p+1}}.$$

Answer 2

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

Answer 3

HINT:

$$\dfrac1n\sum_{r=0}^n\dfrac{(2r+1)^p}{n^p}=\dfrac{(2n+1)^p}{n^{p+1}}+\dfrac1n\sum_{r=1}^n\dfrac{(2r-1)^p}{n^p}$$

Now $$\dfrac1n\sum_{r=1}^n\dfrac{(2r-1)^p}{n^p}=\underbrace{\dfrac1n\sum_{r=1}^{2n}\left(\dfrac rn\right)^p}_{(1)}-\underbrace{\dfrac1n\sum_{r=1}^n\left(\dfrac{2r}n\right)^p}_{(2)}$$

For $(1)$ see See also : Find $\lim\limits_{n \to \infty} \frac{1}{n}\sum\limits^{2n}_{r =1} \frac{r}{\sqrt{n^2+r^2}}$

Now for $(2)$ use $$\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx$$

Answer 4

Just out of personal curiosity.

For the fun of it, I tried to find an approximation of the partial sums $$S_n^{(p)}=\frac{1^p+3^p+\ldots+(2n+1)^p}{n^{p+1}}$$ and found the "simple" result $$\frac{2^p}{p+1}\left(1+\frac {p+1}n+\frac {11p(p+1)}{24 n^2}+\frac {(p-1)p(p+1) } {8n^3}+\frac{127 (p-2) (p-1) p (p+1)}{5760n^4}+\cdots \right)$$ which looks not too bad aven for small values of $n$. In the table, are given the decimal representation of $S_{10}^{(p)}$ for afew values of $p$. $$\left( \begin{array}{ccc} p & \text{exact} & \text{approximation} \\ 1 & 1.209166667 & 1.210000000 \\ 2 & 1.771000000 & 1.771000000 \\ 3 & 2.916105833 & 2.916100000 \\ 4 & 5.118180000 & 5.118190000 \\ 5 & 9.350900000 & 9.351001000 \\ 6 & 17.55979048 & 17.56040110 \\ 7 & 33.63793333 & 33.64080841 \\ 8 & 65.41276444 & 65.42436878 \\ 9 & 128.6969600 & 128.7391174 \\ 10 & 255.5698424 & 255.7116653 \end{array} \right)$$

Answer 5

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\lim_{n \to \infty}{1^{p} + 3^{p} + \cdots + \pars{2n + 1}^{\,p} \over n^{p + 1}} = \lim_{n \to \infty}{\sum_{k = 0}^{n}\pars{2k + 1}^{\,p} \over n^{p + 1}} \\[5mm] = &\ \lim_{n \to \infty}{\sum_{k = 0}^{n + 1}\pars{2k + 1}^{\,p} - \sum_{k = 0}^{n}\pars{2k + 1}^{\,p} \over \pars{n + 1}^{p + 1} - n^{p + 1}} \qquad\pars{~Stolz-Ces\grave{a}ro\ Theorem~} \\[5mm] = &\ \lim_{n \to \infty}{\pars{2n + 3}^{\,p} \over n^{p + 1} \bracks{\pars{1 + 1/n}^{p + 1} - 1}} = 2^{p}\lim_{n \to \infty}{\bracks{1 + 3/\pars{2n}}^{\,p} \over n \bracks{\pars{1 + 1/n}^{p + 1} - 1}}\label{1}\tag{1} \end{align}

Note that $\ds{{\bracks{1 + 3/\pars{2n}}^{\,p} \over n \bracks{\pars{1 + 1/n}^{p + 1} - 1}} \,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\, {1 + 3p/\pars{2n} + 9p\pars{p - 1}/\pars{8n^{2}}\over n\bracks{\pars{p + 1}/n + \pars{p + 1}p/\pars{2n^{2}}}}}$

\begin{align} &\mbox{such that}\quad\lim_{n \to \infty}{\bracks{1 + 3/\pars{2n}}^{\,p} \over n \bracks{\pars{1 + 1/n}^{p + 1} - 1}} = {1 \over p + 1} \\[5mm] &\ \mbox{and}\quad\pars{~\mrm{see}\ \eqref{1}~}\quad \bbx{\lim_{n \to \infty}{1^{p} + 3^{p} + \cdots + \pars{2n + 1}^{\,p} \over n^{p + 1}} = {2^{p} \over p + 1}} \end{align}

Answer 6

This is another path of solution, per request of Claude Leibovici.

If we use the Faulhaber-formula $F_p(n)$ for the polynomial expressions of the sums-of-like powers-problem $S_p(n) = 1^p+2^p + ... + n^p $ and the according polynomial expression $G_p(n)$ for the alternating sum of the same type $A_p(n)$ then of course $$ H_p(n) = {F_p(n)+G_p(n)\over 2 } $$ gives us the sums of like-powers of odd terms only. So the formula in question should be expressed by $$ M_p(n) = {H_p(2n+1) \over n^{p+1} } $$ Here are the first few polynomials $M_p(n)$ as given by Pari/GP:

  M_0(n)=   (n+1)/n                                             
  M_1(n)=   (n^2+2*n+1)/n^2
  M_2(n)=   (4*n^3+12*n^2+11*n+3)/(3*n^3)
  M_3(n)=   (2*n^4+8*n^3+11*n^2+6*n+1)/n^4
  M_4(n)=   (48*n^5+240*n^4+440*n^3+360*n^2+127*n+15)/(15*n^5)

Expanded and with the coefficients reordered we get that list of coefficients

$$\small \begin{array} {} & \cdot 1& \cdot \frac1{n} & \cdot \frac1{n^2} & \cdot \frac1{n^3} & \cdot \frac1{n^4} & \cdot \frac1{n^5} & \cdot \frac1{n^6} & \cdots \\ \hline M_0(n):& 1 & 1 & . & . & . & . & . & \cdots\\ M_1(n):& 1 & 2 & 1 & . & . & . & . & \cdots\\ M_2(n):& 4/3 & 4 & 11/3 & 1 & . & . & . & \cdots\\ M_3(n):& 2 & 8 & 11 & 6 & 1 & . & . & \cdots\\ M_4(n):& 16/5 & 16 & 88/3 & 24 & 127/15 & 1 & . & \cdots\\ M_5(n):& 16/3 & 32 & 220/3 & 80 & 127/3 & 10 & 1& \cdots \\ \vdots &\vdots &\vdots &\vdots &\vdots &\vdots &\vdots &\vdots &\ddots \end{array} $$ Pari/GP gives the fractions always cancelled. If we introduce meaningful cofactors we can write $$\small \begin{array} {} &\cdot 1 &\cdot {1 \over n} &\cdot {11 \cdot 1 \over 24 n^2 } &\cdot {1 \over 4n^3 } &\cdot {127 \over 15 \cdot 64 n^4 } &\cdot {1 \over 16 n^5}& \cdots \\ \hline M_0(n):&1 /1&1 \cdot 1&.&.&.&.&\cdots \\ M_1(n):&2 /2&1 \cdot 2&* \cdot 2&.&.&.&\cdots \\ M_2(n):&4 /3&1 \cdot 4&2 \cdot 4&1 \cdot 4&.& .&\cdots \\ M_3(n):&8 /4&1 \cdot 8&3 \cdot 8&3 \cdot 8&* \cdot 8&.&\cdots \\ M_4(n):&16 /5&1 \cdot 16&4 \cdot 16&6 \cdot 16&4 \cdot 16&1 \cdot 16&\cdots \\ M_5(n):&32 /6&1 \cdot 32&5 \cdot 32&10 \cdot 32&10 \cdot 32&5 \cdot 32&\cdots \\ \vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\ddots \end{array} $$ and get a perfect transparent expression for the sequence of coefficients downwards each column. The only uneasy entries are marked by stars, but that can easily be filled in if the coefficients in the leading row are determined.
I heve not yet determined many of them, but we see that they are rational so far and by the generating process of the wole table it should easily be provable that they shall always be rational. (Each second one seems to follow even a trival formula)
Example, for instance for $p=3$ we get $$M_3(n)= 2^3\left( {1 \over 3+1} + {1 \over n}+ \binom{3}{1} {11 \over 24 n^2} + \binom{3}{2} {1 \over 4 n^3} + \binom{3}{3} {1 \over 2^3 n^4}\right) $$ Here the last coefficient given by the star in the table must come out to be the reciprocal of the leading power of $2$ because in the previous table we have the value $1$ at this position.

Of course, in the limit, when $\lim_{n \to \infty}$ and generalized to any $p$ we have $$ \lim_{n \to \infty} M_p(n) = {2^p\over p+1} $$