# Convergence for series failed using “Ratio Test”

$$\sum_{n=1}^{\infty}\frac{1\cdot 3\cdot 5\cdot ...\cdot (2n-1)}{1\cdot 4\cdot 7\cdot ...(3n-2)}$$

Using Ratio test: $$\lim_{n\rightarrow \infty}\frac{\frac{2(n+1)-1}{3(n+1)-2}}{\frac{2n-1}{3n-2}}$$

which equals to : $$\lim_{n \to \infty}\frac{6n^{2}-n-2}{6n^{2}-n-1}$$ the result for latter is q=1. seemingly this is unclear if divergent or convergent.

The answer for this is convergent....

• No, when $n$ increases, there are more factors. Your ratio is incorrect. – Yves Daoust Oct 9 '18 at 13:54

List out your $$a_n$$ clearly.$$a_n =\prod_{i=1}^n \left(\frac{2i-1}{3i-2}\right)$$

$$\lim_{n \to \infty}\frac{a_{n+1}}{a_n}=\lim_{n \to \infty}\frac{2(n+1)-1}{3(n+1)-2}=\frac{2}{3}$$

Hint:

$$\frac{1\cdot3\cdot5\cdots\cdot(2n+1)}{1\cdot3\cdot5\cdots\cdot(2n-1)}=\frac{1\cdot3\cdot5\cdots\cdot(2n-1)\cdot(2n+1)}{1\cdot3\cdot5\cdots\cdot(2n-1)}=2n+1$$

and not

$$\frac{1\cdot3\cdot5\cdots\cdot(2n+1)}{1\cdot3\cdot5\cdots\cdot(2n-1)}=\frac{2n+1}{2n-1}.$$

The ratio test will work if done correctly $$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| = \lim_{n\to\infty} \frac { \left(\! {\large{\frac {1\,\cdot\,3\,\cdot\,5\,\cdots\,(2n-1)\,\cdot\,(2n+1)\!} {1\,\cdot\,4\,\cdot\,7\,\cdots\,(3n-2)\,\cdot\,(3n+1)\!} }} \! \right) } {\left(\!\large{{\frac {1\,\cdot\,3\,\cdot\,5\,\cdots\,(2n-1)} {1\,\cdot\,4\,\cdot\,7\,\cdots\,(3n-2)} }}\!\!\right)} = \lim_{n\to\infty} \frac{2n+1}{3n+1} = \frac{2}{3}$$ But here's another way . . .

Show that the sequence of fractions $$\frac{1}{1},\frac{3}{4},\frac{5}{7},...,\frac{2n-1}{3n-2}$$ is strictly decreasing.

From that, it follows that the $$n$$-th term is bounded above by $$\bigl({\large{\frac{3}{4}}}\bigr)^{n-1}$$.

Finish via a comparison test.

$$a_k=\prod_{k=1}^{n}\frac{2k-1}{3k-2}=\left(\frac{2}{3}\right)^n\cdot\frac{\Gamma\left(\frac{2}{3}\right)^2}{\pi 2^{2/3}}\cdot B\left(n+\tfrac{1}{2},\tfrac{1}{6}\right)\tag{1}$$ implies $$\begin{eqnarray*} \sum_{k\geq 1}a_k &=& \frac{\Gamma\left(\frac{2}{3}\right)^2}{\pi 2^{2/3}}\int_{0}^{1}\sum_{n\geq 1}\left(\frac{2}{3}\right)^n (1-x)^{-5/6} x^{n-1/2}\,dx \\&=&\frac{\Gamma\left(\frac{2}{3}\right)^2}{\pi 2^{2/3}}\int_{0}^{\pi/2}\frac{4(1-\cos^2\theta)}{\cos^{2/3}\theta(1+2\cos^2\theta)}\,d\theta\\&=&\frac{\Gamma\left(\frac{2}{3}\right)^2}{\pi 2^{2/3}}\int_{0}^{1}12\sqrt{1-z^6}\,\frac{dz}{1+2z^6}\tag{2}\end{eqnarray*}$$ hence trivially $$\sum_{k\geq 1}a_k = \tfrac{1}{2}\cdot\phantom{}_2 F_1\left(1,\tfrac{3}{2};\tfrac{5}{3};\tfrac{2}{3}\right) \leq \frac{3}{\pi}2^{4/3}\,\Gamma\left(\tfrac{2}{3}\right)^2.\tag{3}$$ In general, $$\phantom{}_{p+1}F_p(\ldots; z)$$ is convergent for any $$|z|<1$$.