the remainder of Maclaurin Series

Question

I don't know how to prove/disprove this:

$f(x)$ is a differentiable function from any degree in R. let $R_(n)$ be the remainder of Maclaurin Series of the function f(x).

I need to prove or disprove: if $\lim _{X→0} {\frac{R_n(x)}{x^n}} = 0$ for all $n$ then the radius of convergence of $\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n$ is bigger than 0.5.

Answer 1

There are two problems with the question.

1. $\lim_{n\to\infty}\frac{R_n(x)}{x^n}$ doesn't depend on $n$. Therefore, the part 'for all $n$' is not saying anything. Should it be 'for all $x$'?
2. The series $\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}$ is a series that either converges or not. There is no parameter of which it depends. Should it be $\sum_{n=0}^{\infty}\frac{f^{(n)}(0)x^n}{n!}$?

Assuming the shoulds above.

Putting $x=1$ in $\lim_{n\to\infty}\frac{R_n(x)}{x^n}=0$ we get that $\lim_{n\to\infty}R_n(1)$. Therefore, by definition of $R_n(x)$, $\sum_{n=0}^{\infty}\frac{f^{(n)}(0)1^n}{n!}$ converges (to $f(1)$). Since $\sum_{n=0}^{\infty}\frac{f^{(n)}(0)x^n}{n!}$ is a power series, it must converge absolutely for every $|x|<1$. Therefore the radius of convergence is at least $1>0.5$.

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For the edited question look at $f(x)=\frac{1}{1+3^2x^x}$. This has any number of derivatives at all real $x$. Therefore, by Taylor theorem with Peano's remainder we get that $\lim_{x\to0}\frac{R_n(x)}{x^n}=0$ for all $n$.

On the other hand $\sum_{n=0}^{\infty}\frac{f^{(n)}(0)x^n}{n!}=\sum_{n=0}^{\infty}(-1)^n3^{2n}x^{2n}$ has radius of convergence $1/3<0.5$.