I need to construct a divergent sequence $\displaystyle \{ a_n \}$ such that $\displaystyle a_n>0, \forall n \in \mathbb{N}$, and $\displaystyle \lim \limits_{n \to \infty } \frac{a_{n+1}}{a_n}=1$
I can't think of one
I need to construct a divergent sequence $\displaystyle \{ a_n \}$ such that $\displaystyle a_n>0, \forall n \in \mathbb{N}$, and $\displaystyle \lim \limits_{n \to \infty } \frac{a_{n+1}}{a_n}=1$
I can't think of one
The sequence $\displaystyle a_n=\frac 1n$ converges to 0, because $\displaystyle \lim \limits_{n \to \infty} a_n=\lim \limits_{n \to \infty} \frac{1}{n}=0$.
That will work for the Harmonic series $\displaystyle \sum ^\infty _{n=1} \frac{1}{n}$, but I don't think that will work in this case.
I need something like $\displaystyle \lim \limits_{n \to \infty} a_n=\infty$ and $\displaystyle \lim \limits_{n \to \infty} \frac{a_{n+1}}{a_n}=1$
Please excuse me, I thought we were working on series
So take $\displaystyle a_n=n$I need something like $\displaystyle \lim \limits_{n \to \infty} a_n=\infty$ and $\displaystyle \lim \limits_{n \to \infty} \frac{a_{n+1}}{a_n}=1$
$\displaystyle \frac{a_{n+1}}{a_n}=\frac{n+1}{n}=1+\frac 1n$ which goes to 1 as n goes to infinity.