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?
 A: 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 $f(1) = \pi/2$.
A: Ratio test is not enough.
Example 1:
$$S=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\cdots$$
$$\dfrac{u_n}{u_{n+1}}=\dfrac{\left(\dfrac{1}{n}\right)}{\left(\dfrac{1}{n+1}\right)}=\frac{n+1}{n}=1+\frac{1}{n}$$
Ratio will be $1$ if $n--->\infty$
But we know that S series is divergent.
So the ratio test says us If the ratio is 1.  maybe the series will be convergent. We need more test to determine the result.
I will give you a method how you can approach such series.
$$f(x)=x+\dfrac{1}{2}\dfrac{x^3}{3}+\dfrac{1\cdot 3}{2\cdot 4}\dfrac{x^5}{5}+\dfrac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\dfrac{x^7}{7}+\cdots$$ 
$$f'(x)=1+\dfrac{1}{2}x^2+\dfrac{1\cdot 3}{2\cdot 4}x^4+\dfrac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}x^6+\cdots$$
$$f''(x)=x+\dfrac{1\cdot 3}{2}x^3+\dfrac{1\cdot 3}{2\cdot 4}x^5+\cdots$$ 
$$\int_0^x \frac{f''(x)}{x} dx=x+\dfrac{1}{2}x^3+\dfrac{1\cdot 3}{2\cdot 4}x^5+\cdots$$
$$\int_0^x \frac{1}{x}\int_0^x \frac{f''(t)}{t} dt dx=f(x)$$
$$\int_0^x \frac{f''(x)}{x}dx=xf'(x)$$
$$ \frac{f''(x)}{x}=xf''(x)+f'(x)$$
$$ \int_0^x \frac{f''(x)}{f'(x)} dx=\int_0^x\frac{x}{1-x^2} dx$$
$$ \ln f'(x)=-\frac{1}{2}\ln{(1-x^2)} $$
$$  f'(x)=\frac{1}{\sqrt{1-x^2}} $$
$$  f(x)=\arcsin(x) $$
$$\arcsin(x)=x+\dfrac{1}{2}\dfrac{x^3}{3}+\dfrac{1\cdot 3}{2\cdot 4}\dfrac{x^5}{5}+\dfrac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\dfrac{x^7}{7}+\cdots$$
$x=1$
$$\arcsin(1)=1+\dfrac{1}{2}\dfrac{1}{3}+\dfrac{1\cdot 3}{2\cdot 4}\dfrac{1}{5}+\dfrac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\dfrac{1}{7}+\cdots$$
$$\dfrac{\pi}{2}=1+\dfrac{1}{2}\dfrac{1}{3}+\dfrac{1\cdot 3}{2\cdot 4}\dfrac{1}{5}+\dfrac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\dfrac{1}{7}+\cdots$$
You can use the same method and to show the series $$S=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\cdots$$ is divergent.
$$f(x)=x+\dfrac{x^2}{2}+\dfrac{x^3}{3}+\dfrac{x^4}{4}+\cdots$$ 
$$f'(x)=1+x+x^2+x^3+x^4\cdots=\frac{1}{1-x}$$ 
$$\int_0^x f'(x) dx=f(x)=-\ln{(1-x)}$$
$x=1$
$$f(1)=-\ln{0}=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\cdots$$ 
