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

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.

• This looks like a Riemann sum, in particular if you remove the $2n+1$ term (and $\frac{(2n+1)^p}{n^{p+1}}\to 0$) it is a Riemann sum for $\int_{0}^{1}(2x)^p\,dx$, I believe. – Thomas Andrews Sep 23 '17 at 15:31

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

• Lol, thanks! I like it since it's a continuation of my work and not a brand new solution. I hope that someday I can do proofs like that on my own. Can you give me a bit more details on the small-o notation, best if you can send me a link or recommend a book, somewhere I can understand why $o(1) \to 0$ as $n \to \infty$. – Nikola Sep 23 '17 at 15:53
• @Nikola This is the "little-o notation" See en.wikipedia.org/wiki/Big_O_notation#Little-o_notation – Robert Z Sep 23 '17 at 16:01
• I still don't understand why o(1) goes to 0. Isn't $o(f(x))$ supposed to mean the class of functions which tend to $0$ the same speed as $f(x)$, so how does a constant function approach 0? I know that the $o(\frac{1}{n})$ class of functions tend to 0 but I can't seem to get the case for $o(1)$. – Nikola Sep 24 '17 at 7:28
• @Nikola Not the same speed. this is LITTLE-O. For infinitesimals $g\in o(f(x))$ iff $f(x)/g(x)\to 0$. So $g$ goes faster that $f$. – Robert Z Sep 24 '17 at 7:32
• Aha, thanks, I get it now. – Nikola Sep 24 '17 at 7:34

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

• You beat me to this way forward. – Mark Viola Sep 23 '17 at 15:30
• Which isn't accurate for $p<1$, of course - the limit diverges in that case. – Thomas Andrews Sep 23 '17 at 15:41
• @ThomasAndrews: Isn't the turning point $p=\color{red}{-1}$, for which $x^p$ ceases to be a integrable function over $(0,1)$? – Jack D'Aurizio Sep 23 '17 at 15:43
• Sure, but fpr $p=-1$, the formula is true, since $\frac{2^p}{p+1}$ is undefined there. For $p<-1$ the RHS gives a value, but the value is negative, which can't be right. – Thomas Andrews Sep 23 '17 at 15:44
• Oh, I missed the typo in my first comment. :) I meant $p<-1$. – Thomas Andrews Sep 23 '17 at 16:01

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

• Isn't $(1)$ convered in the your general rule for (2) using $\frac{r}{n}=\frac{2r}{2n}$ so you get $$\frac{2}{2n}\sum_{1}^{2n}f\left(\frac{2r}{2n}\right)$$ has a limit of $$2\int_{0}^{1} f(2x)\,dx$$ – Thomas Andrews Sep 23 '17 at 15:36
• @ThomasAndrews, That's what has been proved in the link – lab bhattacharjee Sep 23 '17 at 15:39
• Yeah, but you don't need the link, just the same thing you use for $(2)$. – Thomas Andrews Sep 23 '17 at 15:39
• It's definitely not a gemoetric series, @Isham – Thomas Andrews Sep 23 '17 at 16:01

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 $p$. $$\left( \begin{array}{ccc} p & \text{exact} & \text{approximation} \\ 1 & 1.209166667 & 1.210000000 \\ 2 & 1.771000000 & 1.771000000 \\ 3 & 2.916105833 & 2.916100000 \\ 4 & 5.118180000 & 5.118190000 \\ 5 & 9.350900000 & 9.351001000 \\ 6 & 17.55979048 & 17.56040110 \\ 7 & 33.63793333 & 33.64080841 \\ 8 & 65.41276444 & 65.42436878 \\ 9 & 128.6969600 & 128.7391174 \\ 10 & 255.5698424 & 255.7116653 \end{array} \right)$$

• Very nice! Seeing the constant $11$ and denominator of $4!$ - are there possibly the Stirling-numbers first kind involved? I've foud a simular but so far only simpler result using the Faulhaber-matrix. But I've not yet taken it farther... – Gottfried Helms Sep 24 '17 at 19:12
• @GottfriedHelms. I added more terms. – Claude Leibovici Sep 25 '17 at 5:35
• In case you are doing this by numerical approximations: Using the Faulhaber-formula for the sum-of-like-powers (say $F_p(n)$) combined with its equivalent for the alternating sums (say $G_p(n)$) evaluating $(F_p(2n+1)+G_p(2n+1))/2$ and the appropriate denominator $n^{p+1}$ one finds polynomials with coefficients which are automatically exact rational values (however I've not yet a general term in this). – Gottfried Helms Sep 25 '17 at 9:56
• @GottfriedHelms. This could be interesting to see. PLease, post a solution for it. I did not perform any numerical approximations. All of that has been done with Taylor series for infinitely large values of $n$ – Claude Leibovici Sep 25 '17 at 10:30
• Claude - see my answer. (I'll be out for the next few hours so can only proceed later if needed) – Gottfried Helms Sep 25 '17 at 12:59


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}

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 $$\lim_{n \to \infty} M_p(n) = {2^p\over p+1}$$