Prove that if x_n --> infinity, then the sequence given by
x_n / x_n+1 is convergent.
I am supposed to use the definition of divergence, so I know there is a number n for which all x_n > M for any M chosen.
to support what i have written in red:
define $\displaystyle x_n$ as $\displaystyle x_{2k}=0, x_{2k+1}=1 \, \forall k \in \mathbb{Z}^+$.
then $\displaystyle x_n$ is divergent since its not convergent. But if we take $\displaystyle M=5$(say) we can't find any $\displaystyle n$ which would satisfy $\displaystyle x_n>M$.
Also can you please explicitly write what definition of $\displaystyle x_n \rightarrow \infty$ you are using??
Consider the sequence $\displaystyle x_n = \left\{ \begin{gathered} n,\text{ n odd} \hfill \\ 2^n ,\text{ n even} \hfill \\ \end{gathered} \right.$.
Is it true that $\displaystyle \left( {x_n } \right) \to \infty ~?$
What can you say about $\displaystyle \frac{{x_n }}{{x_{n + 1} }}~?$