show that $f (x)$ is defined for all values of x

Question

Sea $f(x)=∑_{n=1} ^∞$ $\frac{x^n}{n^n}$ Show that

a) $f(x)$ is defined for all values ​​of x.

b) evaluate the approximate form, where necessary $f(0),f(1),f'(0),f'(1),f''(0).$

c) obtain the MacLaurin series for $f'(x),f''(x)$.

When applying the root criterion, would it look like this? lim $x_{→∞}$ $\sqrt[n]a_n=c$→$ $ if $c>1$ divergent, if , if $ c<1$ convergent, if$ c=1$,gives no information. lim $x_{→∞}$ $\sqrt[n]\frac{x^n}{n^n}$ = lim $x_{→∞}$ $\sqrt[n](\frac{x}{n})^n =$ lim $x_{→∞}\frac{x}{n}=0 $ converges, Help me I don't know what else to do !! I need to finish it

Answer 1

The radius $R$ of convergence of a power series $\sum_n a_n x^n$ is obtained from $$ \frac 1R=\limsup\sqrt[n]{|a_n|}=\limsup\sqrt[n]{\frac 1{n^n}}=\limsup \frac 1n=0$$ and hence is here infinite. Hence the series converges for all $x\in \Bbb C$ and converges absolutely, and the derivative is obtained by summand-wise derivative. That is, $$ f'(x)=\sum_{n=1}^\infty\frac{nx^{n-1}}{n^n}=\sum_{n=0}^\infty \frac{x^n}{(n+1)^n}$$ and by repeating $$ f''(x)=\sum_{n=1}^\infty\frac{nx^{n-1}}{(n+2)^n}=\sum_{n=0}^\infty \frac{(n+1)x^n}{(n+2)^{n+1}.}$$

Clearly $f(0)=0$ as all summands are $0$ and $f'(0)=\frac1{(0+1)^0}=1$ and $f''(0)=\frac{0+1}{(0+2)^{0+1}}=\frac 12$. For $f(1)$ and $f'(1)$, use the first few summands of the series: The summands become smaller very rapidly and the error term is quite small:

$$ f(1)-\sum_{n=1}^N \frac {1}{n^n}=\sum_{n=N+1}^\infty \frac 1{n^n}\le \sum_{n=N+1}^\infty \frac 1{(N+1)^n}=\frac 1{N(N+1)^N}$$ and there is a similar estimate about $f'(x)$.