What will be the $n^{th}$ term of this series?

Question

Original question is to check whether the given series is convergent or divergent.

$$1+\frac{1^2\times 2^2}{1\times 3\times5}+\frac{1^2\times2^2\times3^2}{1\times3\times5\times7\times9}+\cdots$$

But to get to that question, I need to find the $n^{th}$ term of this series. I have figured out that numerator is $(n!)^2$ but can't figure out the denominator. Please help

Answer 1

We have

$$\frac{1}{1 \times 3 \times \ldots \times 9}=\frac{2 \times 4 \times \ldots \times 10}{10!}=\frac{(5!)(2^5)}{10!}=\frac{(2\times 3 -1)!(2^5)}{(2(2\times 3-1))!}$$

In general $$a_n = \frac{(2n-1)!2^{2n-1}(n!)^2}{(2(2n-1))!}$$

Answer 2

My first step in this type of problem is to get rid of the "..." and write it in terms of $\sum$ and $\prod$.

For $1+\frac{1^2\times 2^2}{1\times 3\times5}+\frac{1^2\times2^2\times3^2}{1\times3\times5\times7\times9}+\cdots $ the $n$-th term is $a_n =\dfrac{\prod_{i=1}^n i^2}{\prod_{j=1}^{2n-1} (2i-1)} $.

At this point, there are a few ways to see what happens. I'll start with the ratio of consecutive terms.

$\begin{array}\\ \dfrac{a_{n+1}}{a_n} &=\dfrac{\dfrac{\prod_{i=1}^{n+1} i^2}{\prod_{j=1}^{2n+1} (2i-1)}}{\dfrac{\prod_{i=1}^n i^2}{\prod_{j=1}^{2n-1} (2i-1)}}\\ &=\dfrac{(n+1)^2}{\prod_{j={2n}}^{2n+1} (2i-1)}\\ &=\dfrac{(n+1)^2}{(4n-1)(4n+1)}\\ &\to \dfrac1{16} \qquad\text{for large } n \quad\text{this, as pointed out by Claude Leibovici, is 1/16, not 1/4}\\ \end{array} $

Therefore the sum converges.

Answer 3

$$\sum_{n\geq 1}\frac{n!^2}{(4n-3)!!}=\sum_{n\geq 1}\frac{n!^2 2^{2n-1} (2n-1)!}{(4n-2)!}=\sum_{n\geq 1}\frac{4^n(4n-1)}{\binom{2n}{n}\binom{4n}{2n}}$$ can be written as $$ \sum_{n\geq 1}(4n-1)\frac{8n^2}{4^{2n}}\int_{0}^{\pi/2}(\sin\theta)^{2n-1}\,d\theta\int_{0}^{\pi/2}(\sin\phi)^{4n-1}\,d\phi$$ or as $$ 128\int_{0}^{\pi/2}\int_{0}^{\pi/2}\frac{\sin\theta\sin^3\phi\left(48+\sin^2\theta\sin^4\phi\right) \left(16+5\sin^2\theta\sin^4\phi\right)}{(16-\sin^2\theta\sin^4\phi)^4}\,d\theta\,d\phi$$ which clearly is the integral over a square of a non-negative, continuous and bounded function.
It follows that the original series is convergent. It does not look unlikely that by the tangent half-angle substitution the original series, which equals $$\phantom{}_3 F_2\left(1,2,2;\tfrac{3}{4},\tfrac{5}{4};\tfrac{1}{16}\right)$$ has a nice closed form, but I am honestly too lazy to check if it really is the case. Anyway, the radius of convergence of any $\phantom{}_{p+1}F_p$ equals one, hence the convergence of the original series also follows from the last representation by just noticing that $\frac{1}{16}<1$.