Study the convergence of the series

Question

I need to study the convergence of the following series: $$\sum_{n=1}^{\infty}(\frac{1\cdot3\cdot5\cdot\cdots\cdot(2n+1)}{1\cdot4\cdot7\cdot\cdots\cdot(3n+1)}x^n)^2$$ I tried to expand that term and reduce some terms but didn't succeed, I also tried to apply the ratio test but still nothing. Any help is much appreciated.

Answer 1

The ratio test gives you $${\left(\frac{1 \cdot 3\cdot 5\cdot ...\cdot (2n+1)\cdot (2n+3)}{1\cdot 4\cdot 7\cdot ...\cdot (3n+1)\cdot (3n+4)}x^{n+1}\right)^2\over \left(\frac{1\cdot 3\cdot 5\cdot ...\cdot (2n+1)}{1\cdot 4\cdot 7\cdot ...\cdot (3n+1)}x^n\right)^2}=\frac{(2n+3)^2}{(3n+4)^2} x^2$$ and here you need to take the limit of the ratio. Note that the final result also depends on the value of $x$.

Answer 2

The ratio test helps. Note that $${\left(\frac{1*3*5*...*(2n+3)}{1*4*7*...*(3n+4)}x^{n+1}\right)^2\over \left(\frac{1*3*5*...*(2n+1)}{1*4*7*...*(3n+1)}x^n\right)^2}=\left[{2n+3\over 3n+4}\right]^2 x^2$$and the limit goes to $0$ for $x\in \left(-{3\over 2},{3\over 2}\right)$, therefore the series is convergent over $\left(-{3\over 2},{3\over 2}\right)$.