17
$\begingroup$

First of all, I am asking this question entirely out of curiosity. It basically randomly popped out of my mind.

So I am asking for the value of an infinite series.

Let's call it, $\mathcal{R}=\sum_{n=1}^{\infty}\mathcal{R}_n$

Now I will try to explain what are these $\mathcal{R}_i$'s

Let's take a 1-ball of unit length (1-volume).

It's radius will be $r_1=\frac{1}{2}$

So the first term of my series is $\mathcal{R}_1=1$


Now I will turn the length of this line into the circumference of a circle or a $2$-ball.

It's radius will be $r_2$, you can just do a little calculation and

$r_2=\frac{1}{2\pi}$

And the area of this circle would be, $\pi r_2^2=\frac{1}{4\pi}$

This will be $\mathcal{R}_2=\frac{1}{4\pi}$


Now let's turn the area of this circle into the surface area of a sphere or a $3$-ball,

It's radius would be $r_3$

Again, $r_3=\frac{1}{4\pi}$

And the volume of this sphere will be,

$\frac{4}{3}\pi r_3^3=\frac{1}{48\pi^2}$

This will be, $\mathcal{R}_3=\frac{1}{48\pi^2}$


In general we take an $n$-ball whose radius is $r_n$, then we turn the $n$-volume of this ball into the $n$-surface area of an $n+1$-ball whose radius is $r_{n+1}$.

And $\mathcal{R}$ is the sum over the $n$-volumes of these $n$-balls with radius $r_n$.

So we can write,

$\mathcal{R}=\textstyle\displaystyle\sum_{n=1}^{\infty}V_n(r_n)$

From this we can deduce the formula for $r_n$

Agian in general, we take an $n$-ball with radius $r_n$. Then we set it's $n$-volume to equal the $n$-surface area of an $n+1$-ball. And we define the $n+1$-ball's radius to be $r_{n+1}$. From this we can calculate a recurrence relation for $r_n$,

$V_n(r_n)=S_n(r_{n+1})$

$\textstyle\displaystyle{\frac{\pi^\frac{n}{2}}{\Gamma(\frac{n}{2}+1)}r_n^n=\frac{2\sqrt{\pi}\pi^\frac{n}{2}}{\Gamma(\frac{n+1}{2})}r_{n+1}^n}$

$\textstyle\displaystyle{r_{n+1}=r_n\sqrt[n]{\frac{\Gamma(\frac{n+1}{2})}{2\sqrt{\pi}\Gamma(\frac{n+2}{2})}}}$

with the initial condition of $r_1=\frac{1}{2}$

If we keep expanding $r_n$ using the formula then we get,

$\textstyle\displaystyle{r_n=r_1\prod_{k=1}^{n-1}\sqrt[k]{\frac{\Gamma(\frac{k+1}{2})}{2\sqrt{\pi}\Gamma(\frac{k+2}{2})}}}$

$\textstyle\displaystyle{r_n=\frac{1}{2(2\sqrt{\pi})^{H_{n-1}}}\prod_{k=1}^{n-1}\sqrt[k]{\frac{\Gamma(\frac{k+1}{2})}{\Gamma(\frac{k+2}{2})}}}$

So finally,

$\textstyle\displaystyle{V_n(r_n)=\frac{\pi^\frac{n}{2}}{\Gamma(\frac{n}{2}+1)}r_n^n}$

$\textstyle\displaystyle{\begin{align}\mathcal{R}_n=&\frac{\sqrt{\pi}^{n(1-H_{n-1})}}{2^{n(1+H_{n-1})}\Gamma(\frac{n+1}{2})}\prod_{k=1}^{n}\left(\frac{\Gamma(\frac{k+1}{2})}{\Gamma(\frac{k+2}{2})}\right)^\frac{n}{k}\end{align}}$

So,

$\textstyle\displaystyle{\mathcal{R}=\sum_{n=1}^{\infty}\frac{\sqrt{\pi}^{n(1-H_{n-1})}}{2^{n(1+H_{n-1})}\Gamma(\frac{n+1}{2})}\prod_{k=1}^{n}\left(\frac{\Gamma(\frac{k+1}{2})}{\Gamma(\frac{k+2}{2})}\right)^\frac{n}{k}}$

Here $H_0=0$

$\mathcal{R}\approx 1.0817135\dots$

$\underline{\text{My Question}:-}$

Is there a closed form for $\mathcal{R}$? Or maybe we can write $\mathcal{R}$ in terms of some advance function?

Are there some interesting properties of $\mathcal{R}$?

A list of the first few terms of $\mathcal{R}$:-

$\{\mathcal{R}_n\}_{n=1}^{\infty}$

$=\{$$\mathcal{R}_1$,$\mathcal{R}_2$,$\mathcal{R}_3$,$\mathcal{R}_4$,$\mathcal{R}_5$,$\mathcal{R}_6$,$\mathcal{R}_7$,$\mathcal{R}_8$,$\mathcal{R}_9$,$\mathcal{R}_{10}$,$\dots$$\}$

$\textstyle\displaystyle{=\left\{1, \frac{1}{2^2\pi}, \frac{1}{2^43\pi^2}, \frac{1}{2^\frac{23}{3}3^\frac{4}{3}\pi^\frac{10}{3}}, \frac{1}{2^\frac{31}{3}3^\frac{17}{12}5\pi^\frac{14}{3}}, \frac{1}{2^\frac{72}{5}3^\frac{27}{10}5^\frac{6}{5}\pi^\frac{31}{5}}, \frac{1}{2^\frac{163}{10}3^\frac{179}{60}5^\frac{37}{30}7\pi^\frac{116}{15}}, \cdots\right\}}$

You can yourself find more terms by just plugging the formula in Wolfram alpha.

Notice the terms at the denominator have a very weird property, they are all prime numbers to the power of some rational numbers which is really weird. The more peculiar part is that they are not random prime numbers, but the first $n$ prime numbers.

So there will definitely be some involvement of prime numbers in the evaluation of $\mathcal{R}$.

Observing some patterns:-

Notice that every $n^\text{th}$ $2$ terms namely the ${2n-1}^\text{st}$ and ${2n}^\text{th}$ terms have the first $n$ primes in their denominator.

Also notice that when the $n^\text{th}$ prime first appears at the ${2n-1}^\text{st}$ term it is being raised to the first power and at the ${2n}^\text{th}$ term it is being raised to the power of $\frac{p_n+1}{p_n}$. (If we say that the $n^\text{th}$ prime is $p_n$).

I can't prove these observations, but if they are true to infinity then we can write-

$\textstyle\displaystyle{\mathcal{R}=\sum_{n=1}^{\infty}\left(\frac{1}{p_n\pi^{q_{2n-1}}\prod_{k=1}^{n-1}p_k^{r_{k,2n-1}}}+\frac{1}{p_n^{\frac{p_n+1}{p_n}}\pi^{q_{2n}}\prod_{k=1}^{n-1}p_k^{r_{k,2n}}}\right)}$

For some rational sequence $q_n$ and $r_{m,n}$

It seems like $\forall n\in\mathbb{N}, q_{n+1}>q_n$

And the greatest common divisor of the denominators of $r_{m,n}$ and $q_n$ is not $1$, namely

If $\textstyle\displaystyle{r_{m,n}=\frac{s_{m,n}}{t_{m,n}}}$ and $\textstyle\displaystyle{q_n=\frac{a_n}{b_n}}$ such that $\operatorname{gcd}(s_{m,n},t_{m,n},)=\operatorname{gcd}(a_n,b_n)=1$

Then, $\operatorname{gcd}(t_{1,n},\dots,t_{m,n},b_n)\neq 1$ for $n\gt 3$

I don't know if my observation are true to infinity or not, do they even simplify $\mathcal{R}$, I don't know. But it increases my hope for getting a closed form for $\mathcal{R}$.

If I am able to spot more patterns then I will add it to the post.

Working with the formula itself:-

I hadn't really worked with the formula yet. I spent the last 2 days trying to find more patterns, I think I may have found another one but I need work on that a little bit more before I add that to the question. I am sure is the following progress or regress, but whatever.

$\textstyle\displaystyle{\mathcal{R}_n=\frac{\sqrt{\pi}^{n(1-H_{n-1})}}{2^{n(1+H_{n-1})}\Gamma(\frac{n+1}{2})}\prod_{k=1}^{n}\left(\frac{\Gamma(\frac{k+1}{2})}{\Gamma(\frac{k+2}{2})}\right)^\frac{n}{k}}$

$\textstyle\displaystyle{=\frac{\sqrt{\pi}^{n(1-H_{n})}}{2^{n\left(\frac{1}{2}+H_{n-1}\right)}(n-1)!!}\prod_{k=1}^{n}\left(\frac{\Gamma(\frac{k+1}{2})}{\Gamma(\frac{k+2}{2})}\right)^\frac{n}{k}}$

Now I am going to break $\mathcal{R}$ into two infinite sums as following

$\mathcal{R}=E+O$

Where,

$\textstyle\displaystyle{E=\sum_{n=1}^{\infty}\frac{\pi^{n(1-H_{2n})}}{2^{n(1-2H_{2n-1})}(2n-1)!!}\prod_{k=1}^{2n}\left(\frac{\Gamma(\frac{k+1}{2})}{\Gamma(\frac{k+2}{2})}\right)^\frac{2n}{k}}$

$\textstyle\displaystyle{=\sum_{n=1}^{\infty}\frac{(n-1)!\pi^{n(1-H_{2n})}}{4^{nH_{2n}}(2n-1)!}\prod_{k=1}^{2n}\left(\frac{\Gamma(\frac{k+1}{2})}{\Gamma(\frac{k+2}{2})}\right)^\frac{2n}{k}}$

Here i replaced every $n$ with $2n$ And,

$\textstyle\displaystyle{O=\sum_{n=1}^{\infty}\frac{\sqrt{\pi}^{2n(1-H_{2n})+H_{2n-1}}}{2^{(2n-1)\left(\frac{1}{2}+H_{2n-2}\right)}(2n-2)!}\prod_{k=1}^{2n-1}\left(\frac{\Gamma(\frac{k+1}{2})}{\Gamma(\frac{k+2}{2})}\right)^\frac{2n-1}{k}}$

$\textstyle\displaystyle{=\frac{1}{2\sqrt{2}}\sum_{n=1}^{\infty}\frac{\sqrt{\pi}^{2n(1-H_{2n})+H_{2n-1}}}{2^{(2n-1)(1-H_{2n-1})}(n-1)!}\prod_{k=1}^{2n-1}\left(\frac{\Gamma(\frac{k+1}{2})}{\Gamma(\frac{k+2}{2})}\right)^\frac{2n-1}{k}}$

Here i replaced every $n$ with $2n-1$.

Before I do anything with these $E$ and $O$. I want to first take care of the product inside both of them. Let,

$\textstyle\displaystyle{P_{n}=\prod_{k=1}^{2n}\left(\frac{\Gamma(\frac{k+1}{2})}{\Gamma(\frac{k+2}{2})}\right)^\frac{2}{k}}$

$\textstyle\displaystyle{=\prod_{k=1}^{n}\left(\frac{\Gamma(\frac{2k+1}{2})}{\Gamma(\frac{2k+2}{2})}\right)^\frac{2}{2k}\prod_{k=1}^{n}\left(\frac{\Gamma(\frac{2k}{2})}{\Gamma(\frac{2k+1}{2})}\right)^\frac{2}{2k-1}}$

$\textstyle\displaystyle{=\prod_{k=1}^{n}\left(\frac{\Gamma(k+\frac{1}{2})}{k!}\right)^\frac{1}{k}\prod_{k=1}^{n}\left(\frac{(k-1)!}{\Gamma(k+\frac{1}{2})}\right)^\frac{2}{2k-1}}$

$\textstyle\displaystyle{=\prod_{k=1}^{n}\left(\frac{(2k)!\sqrt{\pi}}{4^kk!^2}\right)^\frac{1}{k}\prod_{k=1}^{n}\left(\frac{4^kk!(k-1)!}{(2k)!\sqrt{\pi}}\right)^\frac{2}{2k-1}}$

$\textstyle\displaystyle{=\prod_{k=1}^{n}\left(\frac{\sqrt{\pi}}{4^k}\right)^\frac{1}{k}\prod_{k=1}^{n}\left(\frac{4^k}{\sqrt{\pi}}\right)^\frac{2}{2k-1}\prod_{k=1}^{n}\left(\frac{(2k)!}{k!^2}\right)^\frac{1}{k}\prod_{k=1}^{n}\left(\frac{k!(k-1)!}{(2k)!}\right)^\frac{2}{2k-1}}$

Since, $\sum_{k=1}^{n}\frac{1}{2k-1}=H_{2n}-\frac{1}{2}H_n$ And by Wolfram Alpha we have,

$\prod_{k=1}^{n}16^\frac{k}{2k-1}=2^{2n+\psi(n+\frac{1}{2})-\psi(\frac{1}{2})}$

Also notice, $\frac{k!(k-1)!}{(2k)!}=B(k,k+1)$ and, $\frac{(2k)!}{k!^2}=\frac{1}{kB(k,k+1)}$ And finally again from Wolfram Alpha, $\prod_{k=1}^{n}k^{-\frac{1}{k}}=e^{\gamma_{1}(n+1)-\gamma_1}$

We have-

$\textstyle\displaystyle{P_{n}^n=2^{n(\psi(n+\frac{1}{2})-2+\gamma+\ln(4))}\pi^{n(H_n-H_{2n})}e^{n(\gamma_1(n+1)-\gamma_1)}\prod_{k=1}^{n}B(k,k+1)^\frac{n}{k(2k-1)}}$

So, $\textstyle\displaystyle{E=\sum_{n=1}^{\infty}\frac{(n-1)!}{(2n-1)!}\pi^{n(1+H_n-2H_{2n})}e^{n(\gamma_1(n+1)-\gamma_1)}2^{n(\psi(n+\frac{1}{2})-2(n+1)H_{2n}+\gamma+\ln(4)-2)}\prod_{k=1}^{n}B(k,k+1)^\frac{n}{k(2k-1)}}$

Now replacing $n$ with $n-\frac{1}{2}$ should give us $P_{2n-1}$ which corresponds to the product inside $O$. I am not sure would it work or not because $n-\frac{1}{2}$ won't be an integer anymore. And I don't know what $\prod_{k=1}^{m}$ means when $m\not\in\mathbb{Z}$. So let's leave that idea and re do the process again. Let,

$\textstyle\displaystyle{Q_n=\prod_{k=1}^{2n-1}\left(\frac{\Gamma(\frac{k+1}{2})}{\Gamma(\frac{k+2}{2})}\right)^\frac{1}{k}}$

$\textstyle\displaystyle{=\prod_{k=1}^{n}\left(\frac{\Gamma(\frac{2k}{2})}{\Gamma(\frac{2k+1}{2})}\right)^\frac{1}{2k-1}\prod_{k=1}^{n-1}\left(\frac{\Gamma(\frac{2k+1}{2})}{\Gamma(\frac{2k+2}{2})}\right)^\frac{1}{2k}}$

$\textstyle\displaystyle{=\prod_{k=1}^{n}\left(\frac{4^k}{\sqrt{\pi}}\right)^\frac{1}{2k-1}\prod_{k=1}^{n-1}\left(\frac{\sqrt{\pi}}{4^k}\right)^\frac{1}{2k}\prod_{k=1}^{n}\left(\frac{k!(k-1)!}{(2k)!}\right)^\frac{1}{2k-1}\prod_{k=1}^{n-1}\left(\frac{(2k)!}{k!^2}\right)^\frac{1}{2k}}$

$\textstyle\displaystyle{=\sqrt{\pi}^{H_{n}-H_{2n}+\frac{1}{2n}}2^{\frac{1}{2}\psi(n+\frac{1}{2})+\frac{\gamma}{2}+\ln(2)+1}\frac{\sqrt[n]{n!}}{(2n)!^\frac{1}{2n}}e^{\gamma_1(n+1)-\gamma_1}\prod_{k=1}^{n-1}B(k,k+1)^\frac{1}{2k(2k-1)}}$

We have-

$\textstyle\displaystyle{Q_n^{2n-1}=\sqrt{\pi}^{(2n-1)(H_{n}-H_{2n}+\frac{1}{2n})}2^{(2n-1)(\frac{1}{2}\psi(n+\frac{1}{2})+\frac{\gamma}{2}+\ln(2)+1)}e^{n(\gamma_1(n+1)-\gamma_1)}\frac{\sqrt[n]{n!}^{2n-1}}{(2n)!^\frac{2n-1}{2n}}\prod_{k=1}^{n-1}B(k,k+1)^\frac{2n-1}{2k(2k-1)}}$

So,

$\textstyle\displaystyle{O=\frac{\sqrt{\pi}}{2\sqrt{2}}\sum_{n=1}^{\infty}\frac{1}{(n-1)!}\sqrt{\pi}^{(2n-1)H_n-4nH_{2n}+2n-\frac{1}{4n}}e^{n(\gamma_1(n+1)-\gamma_1)}2^{(2n-1)(\frac{1}{2}\psi(n+\frac{1}{2})+H_{2n-1}+\frac{\gamma}{2}+\ln(2)}\frac{\sqrt[n]{n!}^{2n-1}}{(2n)!^\frac{2n-1}{2n}}\prod_{k=1}^{n-1}B(k,k+1)^\frac{2n-1}{2k(2k-1)}}$

So finally,

$\textstyle\displaystyle{E=\sum_{n=1}^{\infty}\frac{(n-1)!}{(2n-1)!}\pi^{n(1+H_n-2H_{2n})}e^{n(\gamma_1(n+1)-\gamma_1)}2^{n(\psi(n+\frac{1}{2})-2(n+1)H_{2n}+\gamma+\ln(4)-2)}\prod_{k=1}^{n}B(k,k+1)^\frac{n}{k(2k-1)}}$

$\textstyle\displaystyle{O=\frac{\sqrt{\pi}}{2\sqrt{2}}\sum_{n=1}^{\infty}\frac{1}{(n-1)!}\sqrt{\pi}^{(2n-1)H_n-4nH_{2n}+2n-\frac{1}{4n}}e^{n(\gamma_1(n+1)-\gamma_1)}2^{(2n-1)(\frac{1}{2}\psi(n+\frac{1}{2})+H_{2n-1}+\frac{\gamma}{2}+\ln(2)}\frac{\sqrt[n]{n!}^{2n-1}}{(2n)!^\frac{2n-1}{2n}}\prod_{k=1}^{n-1}B(k,k+1)^\frac{2n-1}{2k(2k-1)}}$

The real thing is that was this really a simplification or a complexification.

And please tell me if there are any mistakes.

$\endgroup$
2

1 Answer 1

9
+25
$\begingroup$

This is not an answer.

I am skeptical about a possible closed form of the summation which would converge extremely fast. Looking at the ratio of sucessive terms for $4\leq n \leq 200$, we have (from a quick and dirty nonlinear regression for which $R^2 >0.999999$) $$\log \left(\frac{R_{n+1}}{R_n}\right) \sim \alpha - \beta\,n^\gamma$$ $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ \alpha & 12.6408 & 0.1425 & \{12.3598,12.9218\} \\ \beta & 14.3159 & 0.1265 & \{14.0663,14.5655\} \\ \gamma & 0.14844 & 0.0008 & \{0.14682,0.15007\} \\ \end{array}$$

Notice that $a_{10}=1.61\times 10^{-21}$, $a_{100}=7.87\times 10^{-522}$, $a_{1000}=3.14\times 10^{-9505}$. So, we do not need to add many terms for a more than decent approximation. Adding the first hundred terms $$R=1.08171353471975721634480208063188816421005944622618865601394019148\cdots$$ which is not recognized by inverse symbolic calculators. However, with an error of $1.35\times 10^{-14}$, $R$ is close to the positive root of the quadratic $7678 x^2-992 x-7911=0$. $$

$\endgroup$
1
  • 12
    $\begingroup$ @RounakSarkar. You changed the question so many times that I gave up $\endgroup$ Sep 18, 2021 at 6:04

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .