Radius of convergence for three series

Question

I need to find the radius of convergence of the 3 following series, but there are no solutions, so I don't know if my steps are correct.

1) $\sum_{n=0}^{\infty}x^{n!}$
2) $\sum_{n=0}^{\infty} \frac{1}{1+2^n}x^n$
3) $\sum_{n=0}^{\infty} \frac{1}{1+n3^n}x^n$

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1) For the first, $a_n=1$ iff $\exists k : k!=n, k \in \mathbb{N}$, zero elsewhere. So the radius is given by $\frac{1}{lim \: sup_{n \to \infty}\sqrt[n]{|a_n|}}$ and the limit superior is max{1,1,0,0,0,1,0,0,....}=1

Thus the radius of convergence is 1.
On the border, we have 1 and -1 to an even power (factorial), which always yields 1, and not 0. So it diverges for both endpoints.

2) Here, we use $\rho = \frac{a_n}{a_{n+1}}$ So we get : $\lim_{n \to \infty}(\frac{1+2^{n+1}}{1+2^{n}})=2$ On the border, we have $\frac{2^n}{1+2^n}=\frac{2^n}{2^n}\frac{1}{\frac{1}{2^n}+1}=1$ for $n \to \infty$, so not toward 0, and thus it diverges.
We also have $\frac{(-1)^n(2)^n}{1+2^n}$, which is an alternating serie. According to the alternating serie test, it should be converging toward 0 for $n \to \infty$ which isn't the case here, so it also diverges for $x=-2$

3) Here, we get $\frac{1+3^{n+1}(n+1)}{1+3^nn}= \frac{3^nn}{3^nn}\frac{\frac{1}{3^nn}+\frac{3(n+1)}{n}}{1+ \frac{1}{3^nn}}=\frac{3}{1}=3$ for $n \to \infty$
On the border, we have $\frac{3^n}{1+n3^n}=\frac{1}{\frac{1}{3^n}+n}$ and $\frac{(-1)^n3^n}{1+n3^n}=\frac{(-1)^n}{\frac{1}{3^n}+n}$. Both cases give the harmonic serie, which diverges. So no convergence on the border.

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Are my results correct ? If not, what did I do wrong or forgot ?
Thanks for your help !

Answer 1

For the first

the ratio test gives the limit

$$\lim_{n\to+\infty}\left\lvert \frac{x^{(n+1)!}}{x^{n!}}\right\rvert=\lim_{n\to+\infty} \lvert x\rvert^{n\cdot n!}$$

if $\lvert x\rvert<1$, the limit is zero. if $\lvert x\rvert>1$ the limit is infinity.

the radius is $R=1$.

When you look for the radius only, you do not need the convergence on the border.

Answer 2

$(1)$ $$\displaystyle\sum_{n=0}^{\infty}x^{n!}<\sum_{n=0}^{\infty}x^n$$ and $\displaystyle\sum_{n=0}^{\infty}x^n$ converges for $|x|<1$. So Radius of convergence is $1$

$(2)$ Use D'Alembert's ratio test: $$\displaystyle\lim_{n\to\infty}\left|\dfrac{a_{n+1}x^{n+1}}{a_nx^n}\right|<1\implies\lim_{n\to\infty}\left|\dfrac{1+2^n}{1+2^{n+1}}\right||x|<1 \implies |x|<2$$ Radius of convergence is $2$.

$(3)$ $$\displaystyle\lim_{n\to\infty}\left|\dfrac{a_{n+1}x^{n+1}}{a_nx^n}\right|<1\implies\lim_{n\to\infty}\left|\dfrac{1+n3^n}{1+(n+1)3^{n+1}}\right||x|<1 \implies |x|<3$$ Radius of convergence is $3$

(When you require radius of convergence, convergence of boundary is not necessary)