# Evaluating the ratio ${{a_{n+1}}\over{a_n}}$ in calculating the radius of convergence for a power series

In calculating the radius of convergence for the power series

$$\sum_{n=1}^\infty {{(2n)!}\over(n!)^2}\ x^n$$

By the ratio test, we let

$$a_n = \lvert {{(2n)!}\over(n!)^2}\ x^n \rvert \quad\quad a_{n+1} = \lvert{{(2n+2)!}\over(n+1)!^2}\ x^{n+1} \rvert$$

$${{a_{n+1}}\over{a_n}} = \lvert {(2n+2)!\over(2n)!}{(n!)^2\over(n+1)!^2}\ x \rvert \longrightarrow \boxed{4\lvert x\rvert < 1} \longrightarrow \boxed{\lvert x\rvert<\frac 14}$$

My method for deriving 4 as the limit of convergence was obtained crudely simply by inputting large values for n and calculating manually; I am unsure as to how to derive 4 formally by simplifying the ratio.

Similarly

$$\sum_{n=1}^\infty {(n!)^3\over(3n)!}\ x^n$$

By the ratio test

$$a_n=\lvert {(n!)^3\over(3n)!}\ x^n \rvert \quad\quad a_{n+1} = \lvert {(n+1)!^3\over(3n+3)!}\ x^{n+1} \rvert$$

$${{a_{n+1}}\over{a_n}} = \lvert {(n+1)!^3\over(n!)^3}{(3n)!\over(3n+3)!}\ x \rvert \longrightarrow \boxed{{1\over27}\lvert x\rvert<1} \longrightarrow \boxed{\lvert x\rvert<27}$$

Again, a formal method for deriving ${1\over27}$ would be much appreciated.

• How about using Stirlig's Formula of Gamma function? – Adelafif Aug 25 '15 at 7:11
• As an aside, $\displaystyle\sum_{n=0}^\infty{2n\choose n}~x^n~=~\frac1{\sqrt{1-4x}}$ . See binomial series for more information. – Lucian Aug 25 '15 at 7:28

Remember that $(n+1)!=(n+1)n!$ and so on. Hence \eqalign{ {(2n+2)!\over(2n)!}{(n!)^2\over(n+1)!^2} &=\frac{(2n+2)(2n+1)(2n)!}{(2n)!}\frac{(n!)^2}{(n+1)^2(n!)^2}\cr &=\frac{(2n+2)(2n+1)}{(n+1)^2}\cr &=\frac{2(2n+1)}{n+1}\cr &=\frac{4+\frac2n}{1+\frac1n}\cr} which tends to $4$ as $n\to\infty$. The case involving $(3n)!$ is very similar.
$$\frac{(2n + 2)!}{(2n)!} \frac{(n!)^2}{(n + 1)!} = \frac{(2n + 2)(2n + 1)}{(n + 1)(n + 1)} = \frac{4n^2 + 6n + 2}{n^2 + 2n + 1} = \frac{4 + \frac{6}{n} + \frac{2}{n^2}}{1 + \frac{2}{n} + \frac{1}{n^2}}$$
$$\frac{(n!)^2}{(n+1)!^2} = \left(\frac{n!}{(n+1)!}\right)^2 = \left(\frac{n!}{n!\cdot (n+1)}\right)^2$$ which you can easily simplify...