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.