$\sum_{n=1}^{\infty} \frac{n 4^n}{(2 n-1)^2(4 n+1)(4 n+3)} \frac{\binom{2 n}{n}}{\binom{4 n}{2 n}}=\frac{4(1+\sqrt{2})}{225}$

Question

I want to show that $\sum_{n=1}^{\infty} \frac{n 4^n}{(2 n-1)^2(4 n+1)(4 n+3)} \frac{\binom{2 n}{n}}{\binom{4 n}{2 n}}=\frac{4(1+\sqrt{2})}{225}.$
I tried manipulating the terms in the sum, but that wasn't very useful. I am new to solving these sorts of problems, so I am not aware of any generating functions which can lead to this expression. Maybe this expression is related to this solution, where we taylor expand $\arcsin{x}$ .

Answer 1

You aim to evaluate $$ \sum_{n=1}^{+\infty}a_n $$ where $$ a_n=4^n\frac{\binom{2n}{n}}{\binom{4n}{2n}}\frac{n}{(2n-1)^2(4n+1)(4n+3)}. $$ Start by noticing that $$ \frac{a_{n+1}}{a_n}=4\frac{(2n-1)^2(n+1)}{(4n+7)(4n+5)n}. $$ This is a rational function in $n$, so you aim to study a hypergeometric sum.

The Gosper algorithm allows you to express this sum as a telescoping sum, which will thus be easy to evaluate. More precisely, Maple (using the command gosper with the sumtools package or Sagemath (using the gosper_term command) indicates that $a_n=g_{n+1}-g_n$ with $$ g_n=\frac{(4n+1)(4n+3)(16n^2-15n-4)}{225n}a_n. $$

Note that once the formula for $g_n$ is given by the algorithm, it is easy to verify that it indeed satisfies $g_{n+1}-g_n=a_n$. In other words, the algorithm doesn't act as a black box but provides you with the steps to prove your formula.

You can then compute $$ \sum_{n=1}^Na_n=g_{N+1}-g_1 $$ and particularly $$ \sum_{n=1}^{+\infty}a_n=\lim_{N\to+\infty}g_{N}-g_1. $$ The Stirling formula leads to $$ \lim_{N\to+\infty}g_{N}=\frac{4\sqrt{2}}{225} $$ and $g_1=-4/225$.

References

• Wolfram Koepf, Hypergeometric Summation, 2nd ed. Springer 2014
• Marko Petkovsek, Herbert Wilf and Doron Zeilberger, A=B, https://www2.math.upenn.edu/~wilf/AeqB.html

Answer 2

$$S=\sum_{n=1}^{\infty} \frac{n 4^n}{(2 n-1)^2(4 n+1)(4 n+3)} \frac{\binom{2 n}{n}}{\binom{4 n}{2 n}}$$

Make the change of variable $n=j+1$

$$S=4\sum_{j=0}^{\infty} \frac{(j+1) 4^{j}}{(2j+1)^2(4j+5)(4j+7)} \frac{\binom{2j+2}{j+1}}{\binom{4j+4}{2j+2}}$$

Now from the formula

$$ j+x = x\frac{(x+1)_j}{(x)_j}$$

where $(x)_{j} = x(x+1)(x+2)\cdots(x+j-1)$ is the rising factorial

$$j+1 = \frac{(2)_{j}}{(1)_{j}}, \quad 2j+1 = \frac{\left(\frac{3}{2}\right)_{j}}{\left(\frac{1}{2}\right)_{j}},\quad 4j+5 = 5\frac{\left(\frac{9}{4}\right)_{j}}{\left(\frac{5}{4}\right)_{j}}, \quad 4j+7 = 7\frac{\left(\frac{11}{4}\right)_{j}}{\left(\frac{7}{4}\right)_{j}}$$

Now, from the Vandermonde's theorem

$$(x)_{n+m} = (x)_{n}(x+n)_m$$

and the multiplication theorem:

$$ (x)_{mn} = m^{mn} \prod_{j=0}^{m-1} \left(\frac{x+j}{m}\right)_{n}$$

$$\binom{2j+2}{j+1} = \frac{(2j+2)!}{(j+1)!(j+1)!} = \frac{(1)_{2j+2}}{(1)_{j+1}(1)_{j+1}} = \frac{2(3)_{2j}}{(2)_{j}(2)_{j}} = \frac{2^{2j+1}\left(\frac{3}{2}\right)_j}{(2)_{j}} $$

$$\binom{4j+4}{2j+2} = \frac{(4j+4)!}{(2j+2)!(2j+2)!} = \frac{(1)_{4j+4}}{(1)_{2j+2}(1)_{2j+2}} = \frac{24(5)_{4j}}{4(3)_{2j}(3)_{2j}} = \frac{6\cdot 2^{4j}\left(\frac{5}{4}\right)_j\left(\frac{7}{4}\right)_j }{\left(2\right)_j\left(\frac{3}{2}\right)_j}$$

After cancelling a lot of terms:

$$S=4\sum_{j=0}^{\infty} \frac{(j+1) 4^{j}}{(2j+1)^2(4j+5)(4j+7)} \frac{\binom{2j+2}{j+1}}{\binom{4j+4}{2j+2}} = \frac{8}{210}\sum_{j=0}^{\infty} \frac{ \left(\frac{1}{2}\right)_j\left(\frac{1}{2}\right)_j(2)_j }{\left(\frac{9}{4}\right)_{j}\left(\frac{11}{4}\right)_{j}}\frac{1}{j!} = \frac{8}{210}{}_3F_{2}\left(\frac{1}{2},\frac{1}{2},2; \frac{9}{4},\frac{11}{4};1\right) $$

Now using the Whipple’s sum formula:

$${{}_{3}F_{2}}\left({a,1-a,c\atop d,2c-d+1};1\right)=\frac{\pi\Gamma\left(d% \right)\Gamma\left(2c-d+1\right)2^{1-2c}}{\Gamma\left(c+\frac{1}{2}(a-d+1)% \right)\Gamma\left(c+1-\frac{1}{2}(a+d)\right)\Gamma\left(\frac{1}{2}(a+d)% \right)\Gamma\left(\frac{1}{2}(d-a+1)\right)} \quad \Re c >0 $$

If $\displaystyle a = \frac{1}{2}, c=2, d=\frac{9}{4}$

$$ S = \frac{8}{210}{}_3F_{2}\left(\frac{1}{2},\frac{1}{2},2; \frac{9}{4},\frac{11}{4};1\right) = \frac{\pi}{210}\frac{\Gamma\left(\frac{9}{4}\right)\Gamma\left(\frac{11}{4}\right)}{\Gamma^2\left(\frac{13}{8}\right)\Gamma^2\left(\frac{11}{8}\right)}$$

From the recursion

$$\Gamma(n+x) = (x)_{n}\Gamma(x)$$

$$ S = \frac{\pi}{210}\frac{\Gamma\left(\frac{9}{4}\right)\Gamma\left(\frac{11}{4}\right)}{\Gamma^2\left(\frac{13}{8}\right)\Gamma^2\left(\frac{11}{8}\right)} = \frac{8\pi}{225} \frac{\Gamma\left(\frac{1}{4}\right)\Gamma\left(\frac{3}{4}\right)}{\Gamma^2\left(\frac{5}{8}\right)\Gamma^2\left(\frac{3}{8}\right)}$$

Finally, from the Euler's reflection formula

$$\Gamma(1-x)\Gamma(x) = \frac{\pi}{\sin(\pi x)} \quad x\notin \mathbb{Z}$$

$$ S = \frac{8}{225} \frac{\sin^2\left(\frac{3\pi}{8}\right)}{\sin\left(\frac{\pi}{4}\right)} = \frac{4(1+\sqrt{2})}{225}$$

$$\boxed{S=\sum_{n=1}^{\infty} \frac{n 4^n}{(2 n-1)^2(4 n+1)(4 n+3)} \frac{\binom{2 n}{n}}{\binom{4 n}{2 n}} = \frac{4(1+\sqrt{2})}{225}}$$