# limit of a series - definition by cases

Be $(a_n)_{n \in\ \mathbb{N}}$ sequences in $\mathbb{R} \backslash \{0\}$. Show that: If the the limit $R = \lim\limits_{n \rightarrow \infty}{|\frac{a_n}{a_{n+1}}|}$ exists definite or indefinite (i.e.$\lim\limits_{n \rightarrow \infty}{{x_n = \pm \infty}})$, then $\sum\limits_{n=0}^{\infty}a_nx^n$ converges for all $x \in \mathbb{R}$ with $|x| < R$ and otherwise it diverges for all $x \in \mathbb{R}$ with $|x| > R$.

My idea is, to show this using the ratio test, but i don't really know how to start..

In fact you can use that $\sum b_n$ converges if $$\overline\lim_{n\to \infty} \sqrt[n]{|b_n|}<1$$ (Cauchy criterion). Now $$0\leq \overline \lim_{n\to \infty} \sqrt[n] {|a_nx^n|}=\frac {|x|}{R}<1$$ where $$R=\frac {1}{\lim_{n\to \infty}\sqrt[n] {a_n}}=\lim\limits_{n \rightarrow \infty}{|\frac{a_n}{a_{n+1}}|}$$
Also $\overline \lim=limsup$ and $limsup=lim$ if the limit exists. But in more of the exercises the limit exists,so use it with $lim$ in order not to get confused. If you know the definition of $limsup$ then it's ok.