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I need help for the following task: Let {$a_n$}$_{n\in\mathbb{N}}$ be a sequence of real numbers such that $\lim\limits_{n \rightarrow \infty}|a_n|^{\frac{1}{n}}=\frac{1}{r}$ with $r$ being a positive real number.

Show that the power series $\sum_{k=0}^\infty a_kx^{2k}$, $x \in \mathbb{R}$ has radius of convergence $\sqrt{r}$.

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I mean this is the formula we have to work with. If we have $x^n$, then $a_n=1$ and the radius is $1$. But in this situation I don't get the result that I should have and I don't know what I did wrong.

Gary
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Analysis_Mark
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  • The $a_n$ is the coefficient of $x^{2n}$. The coefficients of the $x^{2n+1}$s are all zero. Thus the radius of convergence is $$ \mathop {\lim }\limits_{n \to + \infty } \frac{1}{{\sqrt[{2n}]{{|a_n| }}}} = \mathop {\lim }\limits_{n \to + \infty } \frac{1}{{\sqrt {\sqrt[n]{{|a_n| }}} }} = \frac{1}{{\sqrt {1/r} }}=\sqrt r. $$ – Gary Feb 16 '22 at 03:43
  • "If we have $x^n$, then $a_n=1$" Why? Why would $a_n$ be equal to $1$? – Gary Feb 16 '22 at 04:32

1 Answers1

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Let us apply the root test to the proposed power series: \begin{align*} \lim_{n\to\infty}|a_{n}x^{2n}|^{1/n} = x^{2}\times\lim_{n\to\infty}|a_{n}|^{1/n} = \frac{x^{2}}{r} < 1 \Rightarrow |x| < \sqrt{r} \end{align*}

and we are done.

Hopefully this helps !

Átila Correia
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