Determine the radius of convergence of given power series.

(a) $\sum_{n=1}^\infty {(n^{1/n} - 1)x^n}$

(b) $\sum_{n=1}^\infty {{n^c \over n!}x^n}$ ($c \in \mathbb R$)

I tried root test for (a) and got

$$0 \le |n^{1/n} - 1|^{1/n} \lt |n^{1/n}|^{1/n} \le n^{1/n}$$

but I'm not sure if I can use this result.

And for (b) I tried ratio test, and found that if $c=1$, the power series clearly converges for all $x$. But I'm stuck in how to solve the rest case of $c$.

Thank you for advance.


(a) Let $f(x)=x^{-x}$. Then $\lim_{x\to0}f(x)=1$ and $f'(x)=-x^{-x}\bigl(1+\log(x)\bigr)$. Therefore, $\lim_{x\to0}\frac{f(x)-1}x=\lim_{x\to0}f'(x)=+\infty$. So$$\lim_{n\in\mathbb N}\frac{\sqrt[n]n-1}{1/n}=+\infty$$and so, if $n\gg1$, $\displaystyle\frac1n<\sqrt[n]n-1<1$. Since the radius of convergence of both series $\sum_{n=0}^\infty\frac{x^n}n$ and $\sum_{n=0}^\infty x^n$ is $1$, the radius of convergence of your series is also $1$.

(b) $\displaystyle\frac{\frac{(n+1)^c}{(n+1)!}}{\frac{n^c}{n!}}=\frac1{n+1}\left(1+\frac1n\right)^c\to0\times1=0$. So, the radius of convergence is $+\infty$.

  • $\begingroup$ I suppose in (a) you meant $$\lim_{x\to\color{red}{0^+}}x^{-x}=1\;\ldots$$ $\endgroup$ – DonAntonio Jun 12 '17 at 12:16
  • $\begingroup$ @DonAntonio Yes. Since $x^x$ is defined only when $x>0$, I did not feel the need of writing $0^+$ instead of simply writing $0$. $\endgroup$ – José Carlos Santos Jun 12 '17 at 12:18
  • $\begingroup$ Thank you so much! This is very good. $\endgroup$ – user432019 Jun 12 '17 at 12:21
  • $\begingroup$ @user432019 I'm glad I could help. $\endgroup$ – José Carlos Santos Jun 12 '17 at 12:37

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