Raabe Test

**hebby** · February 10th 2010, 04:41 PM

Show that if lim(an+1/an)>1 there exists a natural number such that lim(an+1/an)=>1+1/n.

I have been trying this, but I cant seem to get the proof?

Any Ideas? Thanks

**Drexel28** · February 10th 2010, 04:48 PM

Originally Posted by hebby

Show that if lim(an+1/an)>1 there exists a natural number such that lim(an+1/an)=>1+1/n.

I have been trying this, but I cant seem to get the proof?

Any Ideas? Thanks

Forget the limit part. If $\alpha>1\implies \alpha-1>0$ and by the Archimedean principle there exist some $n\in\mathbb{N}$ such that $\frac{1}{n}<\alpha-1\implies \alpha>1+\frac{1}{n}$

**hebby** · February 10th 2010, 04:57 PM

So how does this link back to the lim(an+1/an)>1.....? I can see what you are using.

**Drexel28** · February 10th 2010, 08:04 PM

Originally Posted by hebby

So how does this link back to the lim(an+1/an)>1.....? I can see what you are using.

I'm not sure what you mean here? Let $\alpha=\lim\text{ }\frac{a_{n+1}}{a_n}$ assuming it exists.

**hebby** · February 10th 2010, 08:15 PM

lim(an+1/an)>1 there exists a natural number such that lim (an+1/an)=>1+1/n..........the () are absolute values.....link in the ratio test...i dont know how to write it....sorry...would this change anything

**Drexel28** · February 10th 2010, 08:17 PM

Originally Posted by hebby

lim(an+1/an)>1 there exists a natural number such that lim (an+1/an)=>1+1/n..........the () are absolute values.....link in the ratio test...i dont know how to write it....sorry...would this change anything

In our example $\alpha$ is any number. You tell me, does it make a difference?

**hebby** · February 10th 2010, 08:20 PM

no I dont think so...its just the proof for lim (an+1/an)<1 looked a bit different in my notes, which yields <= 1-a/n for all n >= N....this is for the Raabe test

**Drexel28** · February 10th 2010, 08:24 PM

Originally Posted by hebby

no I dont think so...its just the proof for lim (an+1/an)<1 looked a bit different in my notes, which yields <= 1-a/n for all n >= N....this is for the Raabe test

Raabe test for series convergence?

**hebby** · February 10th 2010, 08:26 PM

yes...the one you use once ur ratio test yields 1...so we try the raabe test for series

**Drexel28** · February 10th 2010, 08:33 PM

Originally Posted by hebby

yes...the one you use once ur ratio test yields 1...so we try the raabe test for series

Ok. I understand what Raabe's test is. I am just wondering if your problem is not what it seems. I literally interpreted your question as if $\lim\text{ }\left|\frac{a_{n+1}}{a_n}\right|>1$ then there exists some $n'\in\mathbb{N}$ such that $\lim\text{ }\left|\frac{a_{n+1}}{a_n}\right|\geqslant 1+\frac{1}{n'}$

**hebby** · February 10th 2010, 08:36 PM

yes this is what I am having a little trouble with...proving this...can u help?

**Drexel28** · February 10th 2010, 08:38 PM

Originally Posted by hebby

yes this is what I am having a little trouble with...proving this...can u help?

That is exactly what I just proved.

**hebby** · February 10th 2010, 08:53 PM

SO we let L=lim (an+1/an)>1

then L>1.....L-1>0...the via the Archimedean principle there exist a natural number N such that L-1>1/n........L=1+1/n......done?

**Drexel28** · February 10th 2010, 09:21 PM

Originally Posted by hebby

SO we let L=lim (an+1/an)>1

then L>1.....L-1>0...the via the Archimedean principle there exist a natural number N such that L-1>1/n........L=1+1/n......done?

Change that last equals sign to a > and yes.

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