1. ## Raabe Test

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

2. 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 $\frac{1}{n}<\alpha-1\implies \alpha>1+\frac{1}{n}$

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

4. 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.

5. 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

6. 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?

7. 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

8. 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?

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

10. 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 $\lim\text{ }\left|\frac{a_{n+1}}{a_n}\right|\geqslant 1+\frac{1}{n'}$

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

12. 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.

13. 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?

14. 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.