$\sum \frac{(-2x)^n}{e^n}$ intervals of convergence Is the interval of convergence of these infinite here?

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*$$\sum \frac{(-2x)^n}{e^n}$$

*$$\sum \frac{nx^n(n+1)}{n!+n}$$
I check it using the root test 1., root test it's $\lim_{n\to\infty}\Big|\frac{(-2x)^n (-2x)}{e^n e} \frac{e^n}{(-2x)^n}\Big|$. So it's cancel out and no limit is here. It becomes $\frac{2|x|}{e}$ here yes? So limit is zero.
2., then this I calculate using the root test to be $\lim_{n\to \infty} \Big|\frac{x (n+2)(n!+n)}{n ((n+1)! +(n+1))}\Big|$. And limit is zero.
To summarize, if limit is zero, then radius of convergence is infinity. So the interval of convergence is infinite. Right?

You seem to have gotten the second one.
For the first,  by Cauchy-Hadamard, we have $r=\limsup_{n\to\infty}\dfrac1{|(-2/e)^n|^{1/n}}=e/2$.
