Why is $\sum_{k=0}^{\infty} \frac{k!}{\prod_{j=0}^{k} \left(2j+3\right)} = 2-\frac{\pi}{2}$ and how is this solution derived

factorialproductssequences-and-series

I recently came across the problem:

$$\sum_{k=0}^{\infty} \frac{k!}{\prod_{j=0}^{k} \left(2j+3\right)}$$ and decided to try finding its solution. I started off by writing a program that gave me the answer $0.4292036732051…$ which I found probably meant the solution was $2-\frac{\pi}{2}$, but I wanted to see if I could prove it. I started by manipulating the denominator into a double factorial resulting in:
$$\sum_{k=0}^{\infty} \frac{k!}{\left(2k+3\right)!!}$$
Then, I thought things would be easier to work with if I only had regular factorials. This gave me:
$$\sum_{k=0}^{\infty} \frac{4\cdot2^k k! \left(k+2\right)!}{\left(2k+4\right)!}$$
I unfortunately had little clue to go from here considering I haven't dealt with factorials often, much less in infinite series such as this one. I am curious how it is possible to go forth from here and also how to solve infinite series that are similarly structured. Any help would be appreciated.

Best Answer

We obtain \begin{align*} \sum_{k=0}^\infty\frac{k!}{(2k+3)!!}&=\sum_{k=0}^\infty\frac{k!(2k+2)!!}{(2k+3)!}\\ &=\sum_{k=0}^\infty\frac{k!2^{k+1}(k+1)!}{(2k+3)!}\\ &=\sum_{k=0}^\infty\binom{2k}{k}^{-1}\frac{2^k}{(2k+1)(2k+3)}\\ &=\sum_{k=0}^\infty\binom{2k}{k}^{-1}\frac{2^{k-1}}{2k+1}-\sum_{k=0}^\infty\binom{2k}{k}^{-1}\frac{2^{k-1}}{2k+3}\tag{1} \end{align*}

We use a representation of reciprocal binomial coefficients via the Beta function:

\begin{align*} \binom{n}{k}^{-1}=(n+1)\int_0^1z^k(1-z)^{n-k}\,dz\tag{2} \end{align*}

and the left-hand series of (1) can be calculated as

\begin{align*} \color{blue}{\sum_{k=0}^\infty\binom{2k}{k}^{-1}\frac{2^{k-1}}{2k+1}} &=\sum_{k=0}^\infty 2^{k-1}\int_0^1z^k(1-z)^k\,dz\tag{3}\\ &=\frac{1}{2}\int_{0}^{1}\sum_{k=0}^\infty \left(2z(1-z)\right)^k\,dz\\ &=\frac{1}{2}\int_{0}^1\frac{dz}{1-2z(1-z)}\tag{4}\\ &=\frac{1}{2}\int_{0}^{1}\frac{dz}{z^2+(1-z)^2}\\ &=\frac{1}{2}\int_{0}^{\infty}\frac{du}{1+u^2}\tag{5}\\ &\,\,\color{blue}{=\frac{\pi}{4}}\tag{6} \end{align*}

Comment:

In (3) we use the identity (2).
In (4) we apply the geometric series expansion.
In (5) we use the substitution $u=\frac{1-z}{z}, du=-\frac{1}{z^2}dz$.

We also want to apply (2) to the right-hand series of (1). To do this conveniently we need some preparatory work: \begin{align*} \sum_{k=0}^\infty&\binom{2k}{k}^{-1}\frac{2^{k-1}}{2k+3}\\ &=\sum_{k=0}^\infty\frac{k!k!}{(2k)!}\cdot\frac{2^{k-1}}{2k+3}\\ &=\sum_{k=0}^\infty\frac{k!(k+1)!(2k+1)}{(2k+1)!(k+1)}\cdot\frac{2^{k-1}}{2k+3}\\ &=\sum_{k=0}^\infty\frac{(k+1)!(k+1)!}{(2k+2)!}\cdot\frac{2^{k+1}}{2k+3}-\sum_{k=0}^\infty\frac{k!(k+1)!}{(2k+1)!(k+1)}\cdot\frac{2^{k-1}}{2k+3}\\ &=\sum_{k=0}^\infty\binom{2k+2}{k+1}^{-1}\frac{2^{k+1}}{2k+3}-\sum_{k=0}^\infty\binom{2k}{k}^{-1}\frac{2^{k-1}}{(2k+1)(2k+3)}\\ &=\sum_{k=0}^\infty\binom{2k+2}{k+1}^{-1}\frac{2^{k+1}}{2k+3} -\sum_{k=0}^\infty\binom{2k}{k}^{-1}\frac{2^{k-2}}{2k+1} +\sum_{k=0}^\infty\binom{2k}{k}^{-1}\frac{2^{k-2}}{2k+3}\tag{7}\\ \end{align*} In the last line (7) we use a partial fraction decomposition as we did in (1).

We are now well prepared to do the calculation. We obtain together with (6):

\begin{align*} \color{blue}{\sum_{k=0}^\infty \binom{2k}{k}^{-1}\frac{2^{k-2}}{2k+3}} &=\sum_{k=0}^\infty\binom{2k+2}{k+1}^{-1}\frac{2^{k+1}}{2k+3}-\frac{\pi}{8}\\ &=\sum_{k=0}^\infty2^{k+1}\int_{0}^1z^{k+1}(1-z)^{k+1}\,dz-\frac{\pi}{8}\\ &=\sum_{k=1}^\infty2^k\int_{0}^1z^k(1-z)^k\,dz-\frac{\pi}{8}\\ &=\frac{\pi}{2}-2^0\int_{0}^1\,dz-\frac{\pi}{8}\\ &\,\,\color{blue}{=\frac{3}{8}\pi-1}\tag{8} \end{align*}

We finally conclude from (1) together with (6) and (8) \begin{align*} \color{blue}{\sum_{k=0}^\infty\frac{k!}{(2k+3)!!}} &=\sum_{k=0}^\infty\binom{2k}{k}^{-1}\frac{2^{k-1}}{2k+1}-\sum_{k=0}^\infty\binom{2k}{k}^{-1}\frac{2^{k-1}}{2k+3}\\ &=\frac{\pi}{4}-2\left(\frac{3}{8}\pi-1\right)\\ &\,\,\color{blue}{=2-\frac{\pi}{2}} \end{align*}

and the claim follows.

Related Solutions

Can a series $\sum_{n=1}^{\infty} (\frac{1}2)^n=1$ be infinitely substituted into itself

What you end up with is the relation

$$x = \sum_{n=1}^\infty \left(\frac{x}{2}\right)^n.$$

If we assume $x$ to be complex we must have $|x| < 2$ or the RHS diverges. Then, summing the geometric series

$$x = -1 + \sum_{n=0}^\infty \left(\frac{x}{2}\right)^n$$ $$x = -1 + \frac{1}{1-\frac{x}{2}}$$ $$-\frac{1}{2}x^2 +\frac{1}{2}x = 0$$

gives $x = 0 \vee x = 1$ as the only solutions.

Simplify $\pi = 2 \sum_{n=0}^{\infty} \frac{\left(\frac{1}{2}\right)_n \left(\frac{1}{2}\right)_n}{\left(\frac{3}{2}\right)_n} \frac{1^n}{n!}$

If all you're looking for is a simplification you can use the definition $(a)_n=\Gamma(a+n)/\Gamma(a)$ and the property $\Gamma(z+1)=z\Gamma(z)$ to write $$ \begin{aligned} \pi &=2\sum_{n=0}^\infty\frac{(1/2)_n (1/2)_n}{(3/2)_n} \frac{1^n}{n!}\\ &=2\frac{\Gamma(3/2)}{\Gamma(1/2)}\sum_{n=0}^\infty\frac{\Gamma(1/2+n) }{\Gamma(3/2+n)} \frac{(1/2)_n}{n!}\\ &=\sum_{n=0}^\infty\frac{1}{1/2+n}\frac{(1/2)_n}{n!}. \end{aligned} $$

If you cannot use the Pochhammer symbol but double factorials are fine note that $$ (1/2)_n=\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{5}{n}\cdots\frac{2n-1}{2}=\frac{1\cdot 3\cdot 5\cdots(2n-1)}{2^n}=\frac{(2n-1)!!}{2^n} $$ So that we have the equivalent form $$ \begin{aligned} \pi &=\sum_{n=0}^\infty\frac{(2n-1)!!}{1/2+n}\frac{2^{-n}}{n!}. \end{aligned} $$

Typing

Sum[1/(1/2 + n) Factorial2[2 n - 1]/n! 1/2^n, {n, 0, Infinity}]

into Mathematica does in fact return the value $\pi$.

Edit:

I do not write in C but chat GPT does. Here is C code to sum up the first eleven terms

#include <stdio.h>
#include <math.h>

double factorial2(int n) {
    double res = 1;
    for (int i = 1; i <= n; i++) {
        res *= (2 * i - 1);
    }
    return res;
}

int main() {
    int n;
    double sum = 0;
    for (n = 0; n <= 10; n++) {
        double term = 1 / ((1.0 / 2) + n) * factorial2(2 * n - 1) / tgamma(n + 1) * pow(0.5, n);
        sum += term;
    }
    printf("%.15f\n", sum);
    return 0;
}

Best Answer

Related Solutions

Can a series $\sum_{n=1}^{\infty} (\frac{1}2)^n=1$ be infinitely substituted into itself

Simplify $\pi = 2 \sum_{n=0}^{\infty} \frac{\left(\frac{1}{2}\right)_n \left(\frac{1}{2}\right)_n}{\left(\frac{3}{2}\right)_n} \frac{1^n}{n!}$

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