Radii of convergence for complex series

I need to find the radii of convergence for these series:

$1. \sum_{n=1}^\infty (2+(-1)^n)^n z^{2n}$

$2. \sum_{n=1}^\infty (n+a^n)z^n, a \in C $

$3. \sum_{n=1}^\infty 2^n z^{n!}$

Starting with the first power series, I attempted this by claiming that $$a_n =(2+(-1)^k)^k$$ if $n = 2k, k\ge 1$, and $0$ otherwise. Then,

$$|a_n|^{1/n} = ((2+(-1)^k)^k)^\frac{1}{2k}= \sqrt{(2+(-1)^k}$$

Therefore,
$$ \limsup|a_n|^{1/n} = \limsup = \sqrt{(2+(-1)^k} = \sqrt3$$

The radius of convergence is $\frac{1}{\sqrt3}$. I would just like to confirm if this approach is feasible, and if not, what I should change about it.

As for the second series, my first idea was to split up the sum like this:
$$\sum_{n=1}^\infty (n+a^n)z^n = \sum_{n=1}^\infty nz^n + \sum_{n=1}^\infty (az)^n $$

Is it acceptable to say that the radius of convergence for this one is the radii of convergences of the two series that were broken up?

Lastly, for the third, I was considering using a ratio test for this one, but I fail to see how that would help me, nor do I see how any other option would be helpful.

Any advice would be much appreciated. Thank you!

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Looks like your answer for the first one is fine.

For the second one, consider the power series $\sum_n (1-1)z^n.$ Obviously the radius of convergence is $\infty.$ But if you split it up, as you are hoping is a legal maneuver, you are left with $\sum_n z^n + \sum_n (-1)z^n.$ The radius of convergence of both of these power series is $1,$ but that’s a long way from being correct for the original power series. So you’ll need to think about $\sum (n+a^n)z^n$ some more.

For the third one, the ratio test works just fine: The ratios of absolute values are

$$\tag 1 2|z|^{(n+1)! -n!} = 2|z|^{nn!}.$$

If $|z|<1,$ $(1)\to 0$ fast, so we have convergence. If $|z|>1,$ $(1) \to \infty$ fast, so we have divergence. That’s enough to nail down the radius of converge: It’s $1.$