Show that the series converges if $ \ |x|<1 \ $

Question

Show that $$\sum_{n=1}^{\infty} \frac{1 ^2 \cdot 3^2 \cdot 5^2 \cdots(2n-1)^2}{2^2 \cdot 4^2 \cdot 6^2 \cdots(2n)^2} x^{2n}$$ converges if $ \ |x|<1 \ $

Answer:

Let

$$u_{2n}=\frac{1 ^2 \cdot 3^2 \cdot 5^2\cdots(2n-1)^2}{2^2 \cdot 4^2 \cdot 6^2 \cdots(2n)^2} x^{2n}, \quad u_{2n+1}= \frac{1 ^2 \cdot 3^2 \cdot 5^2 \cdots(2n+1)^2}{2^2 \cdot 4^2 \cdot 6^2\cdots(2n+2)^2} x^{2n+1}$$

Now,

$$ l=\lim_{n \to \infty} \frac{u_{2n+1}}{u_{2n}}= \lim_{n \to \infty} \frac{1 ^2 \cdot 3^2 \cdot 5^2\cdots(2n+1)^2}{2^2 \cdot 4^2 \cdot 6^2 \cdots(2n+2)^2} x^{2n+1} \times \frac{2^2 \cdot 4^2 \cdot 6^2 \cdots(2n)^2}{1^2 \cdot 3^2 \cdot 5^2 \cdots(2n-1)^2} \frac{1}{x^{2n}}\ =\lim_{n \to \infty} \frac{(2n+1)^2}{(2n+2)^2} x=x$$

Thus,

$|l=x|<1 \Rightarrow |x|<1$ if converges.

But I think $ u_{2n+1}$ is not correct what I wrote.

Help me out.

Answer 1

Globally, the idea is correct, but you should have defined $u_n$ as$$\frac{1^2.3^2\cdots(2n-1)^2}{2^2.4^2\cdots(2n)^2}x^{2n}.$$So$$u_{n+1}=\frac{1^2.3^2\cdots(2n+1)^2}{2^2.4^2\cdots(2n+2)^2}x^{2n+2}$$and$$\lim_{n\to\infty}\left|\frac{u_{n+1}}{u_n}\right|=\lim_{n\to\infty}\frac{(2n+1)^2}{(2n+2)^2}|x|^2=|x|^2.$$Therefore, the series converges when $|x|^2<1(\iff|x|<1)$.