finding the radius of convergence of a sequence could you please look and verify if what i did is correct? i am requested to find the radius of convergence for those two sequences(sums):
1)$\sum _{n=0}^{\infty }\:\frac{n!}{\left(2n\right)!}x^n$
2)$\sum _{n=2}^{\infty }\:\frac{\left(1-x\right)^{5n}}{nln\left(n\right)\cdot 5^n}$
what i did:
1)$\sum _{n=0}^{\infty }\:\frac{n!}{\left(2n\right)!}x^n$, i did the ratio test:$ \lim _{n\to \infty \:}\left(\left|\frac{\frac{\left(n+1\right)!x^{\left(n+1\right)}}{\left(2\left(n+1\right)\right)!}}{\frac{n!x^n}{\left(2n\right)!}}\right|\right) $, to obtain that $\lim _{n\to \infty \:}\left(\left|\frac{x}{2\left(2n+1\right)}\right|\right)$, took $|\frac{x}{2}|$ out, $\left|\frac{x}{2}\right|\cdot \lim \:_{n\to \infty \:}\left(\left|\frac{1}{2n+1}\right|\right)$, so basically $\left|\frac{x}{2}\right|\cdot \lim \:_{n\to \infty \:}\left(\left|\frac{1}{2n+1}\right|\right)%$, which means that $x=0$. since x=0, does it mean that it converges for all x? ($(-\infty,\infty)$)
2)rearrenged, $(\sum _{n=2}^{\infty }\left(1-x\right)^{5n})(\sum _{n=2}^{\infty }\:\frac{1}{nln\left(n\right)\cdot 5^n})$, now it seems that the root test is easier with the root test:$\lim _{n\to \infty }\left(\sqrt[n]{\frac{1}{nln\left(n\right)\cdot \:5^n}}\right)$ = $\lim _{n\to \infty }\left(\frac{1}{nln\left(n\right)\cdot \:5^n}\right)^{\frac{1}{n}}$ =$\frac{1}{1*1*1*5}$=$\frac{1}{5}$, and the second sequence is: $\sum _{n=2}^{\infty \:}\left(1-x\right)^{5n}$, using the root test again: $\lim _{n\to \infty \:}\left(\left|\left(\left(1-x\right)^{5n}\right)^{\frac{1}{n}}\right|\right)$ = $\left|\left(1-x\right)^5\right|\cdot \lim \:_{n\to \infty \:}\left(\left|1\right|\right)$ . now checking $\left|\left(1-x\right)^5\right|<1$, which gives the range (0,2). so what is the range of convergence in this case?
what i did is correct? if i did any mistake, please correct me so i can learn and improve.
thank you very much for your help.
 A: It is indeed a good idea to use the "d'Alembert ratio test". For the first, one has
$$
\left|\frac{\frac{\left(n+1\right)!x^{\left(n+1\right)}}{\left(2\left(n+1\right)\right)!}}{\frac{n!x^n}{\left(2n\right)!}}\right|=\frac{|x|.(n+1)}{(2n+1)(2n+2)}
$$
which tends to $0$ for all $x$, so the radius is infinite. 
For the second one, you cannot rearrange as you do and the ratio is
$$
\Big|\frac{(1-x)^5.5.(n+1)ln(n+1)}{(n)ln(n)}\Big|
$$
which tends to $5|(1-x)|^5$, can you then deduce the radius ? (it is a series "around $1$" however).  
A: In your first series you mentioned " since x=0, does it mean that it converges for all x?  in (−∞,∞)" You should rephrase your sentence as " $$\left|\frac{x}{2}\right|\cdot \lim \:_{n\to \infty \:}\left(\left|\frac{1}{2n+1}\right|\right)=0 $$ for every x, the series converges for $ x\in (-\infty, \infty )$
For your second series you have splitted one series into product of two series which is not allowed. However, you have started with the right step of the Root Test.  
