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\noindent {\bf Question}

\noindent Prove, if $\{ a_n\}$ is a sequence satisfying
$\lim_{n\rightarrow\infty} |a_n|^{1/n} = L\neq 0$, then the power
series $\sum_{n=0}^\infty a_n x^n$ has radius of convergence
$\frac{1}{L}$.
\medskip

\noindent {\bf Answer}

\noindent The condition that the $a_n$ satisfy is similar to the
condition of the root test, and so we apply the root test to the
power series $\sum_{n=0}^\infty a_n x^n$. Namely, we calculate
\[ \lim_{n\rightarrow\infty} \left| a_n x^n\right| ^{1/n} = |x|
\lim_{n\rightarrow\infty} \left| a_n \right| ^{1/n} = L |x|. \]
Hence, the series converges absolutely for $L |x| < 1$, that is
$|x| < \frac{1}{L}$, and diverges for $L|x| > 1$, and so the
radius of convergence of this series is $\frac{1}{L}$, as desired.

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