# [SOLVED] Positive and divergent infinite sequence

• Nov 30th 2008, 04:44 AM
akolman
[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 AM
Moo
Hello,
Quote:

Originally Posted by akolman
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)

$\displaystyle a_n=\frac 1n$ ? (Tongueout)

Note that if you apply the ratio test and find the limit to be 1, the test is inconclusive.
• Nov 30th 2008, 05:41 AM
akolman
Quote:

Originally Posted by Moo
Hello,

$\displaystyle a_n=\frac 1n$ ? (Tongueout)

Note that if you apply the ratio test and find the limit to be 1, the test is inconclusive.

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 AM
Moo
Quote:

Originally Posted by akolman
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.

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 $\displaystyle \lim \limits_{n \to \infty} \frac{a_{n+1}}{a_n}=1$
So take $\displaystyle a_n=n$ :D

$\displaystyle \frac{a_{n+1}}{a_n}=\frac{n+1}{n}=1+\frac 1n$ which goes to 1 as n goes to infinity.