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 (Worried)

Printable View

- Nov 30th 2008, 04:44 AMakolman[SOLVED] Positive and divergent infinite sequence
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 (Worried) - Nov 30th 2008, 05:24 AMMoo
- Nov 30th 2008, 05:41 AMakolman
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 $\displaystyle \lim \limits_{n \to \infty} \frac{a_{n+1}}{a_n}=1$ - Nov 30th 2008, 05:57 AMMoo
Please excuse me, I thought we were working on series :(

Quote:

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.