<s>[INST] Convergence of $1+\frac{1}{2}\frac{1}{3}+\frac{1\cdot 3}{2\cdot 4}\frac{1}{5}+\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\frac{1}{7}+\cdots$ Is it possible to test the convergence of $1+\dfrac{1}{2}\dfrac{1}{3}+\dfrac{1\cdot 3}{2\cdot 4}\dfrac{1}{5}+\dfrac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\dfrac{1}{7}+\cdots$ by Gauss test?
If I remove the first term I can see $\dfrac{u_n}{u_{n+1}}=\dfrac{(2n+2)(2n+3)}{(2n+1)^2}
\\=\dfrac{\left(1+\dfrac{1}{n}\right)\left(1+\dfrac{3}{2n}\right)}{\left(1+\dfrac{1}{2n}\right)^2}
\\={\left(1+\dfrac{1}{n}\right)\left(1+\dfrac{3}{2n}\right)}{\left(1+\dfrac{1}{2n}\right)^{-2}}\\=\left(1+\dfrac{5}{2n}+\dfrac{3}{2n^2}\right)\left(1-\dfrac{1}{n}\ldots\right)\\=1+\dfrac{3}{2n}+O\left(\dfrac{1}{n^2}\right)$
So the series is convergent.
Is it a correcct attempt?
 [/INST] I'm not sure about your result; the ratio test is inconclusive anyway.  The way I look at this, I express each term as
$$b_k = \frac{a_k}{2 k+1}$$
where
$$a_k = \frac{1}{2^{2 k}} \binom{2 k}{k}$$
Using, e.g., Stirling's formula, you may show that
$$a_k \sim \frac{1}{\sqrt{\pi k}} \quad (k \to \infty)$$
so that the ratio test on the coefficients provides
$$\frac{b_{k+1}}{b_k} \sim \left(1+\frac{1}{k}\right)^{-1/2} \left (1+\frac{2}{2 k+1}\right) \sim 1+\frac{1}{2 k}\quad (k \to \infty)$$
The limit test is inconclusive, as the limit if the ratio is $1$.  However, we may see that the series converges by the comparison test because the terms in the sum behave as 
$$b_k \sim \frac{1}{2 \sqrt{\pi}} k^{-3/2} \quad (k \to \infty)$$
That said, we know that the series converges to $\pi/2$ as follows: consider the generating function
$$f(x) = \sum_{k=0}^{\infty} a_k \frac{x^{2 k+1}}{2 k+1}$$
Then for all $x$ within the radius of convergence of the series on the right, we may differentiate to get
$$f'(x) = \sum_{k=0}^{\infty} a_k \, x^{2 k} = \frac{1}{\sqrt{1-x^2}}$$
Therefore
$$f(x) = \arcsin{x}$$
and the series has value $f(1) = \pi/2$.
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