Sunday, 2 March 2014

real analysis - Proof of Raabe's Test for divergence



Question: Prove that if un>0 and limnn(unun+11)=l , then n=1un diverges if l<1.



My Attempt: We choose a positive number ϵ such that l+ϵ<1.




Since limnn(unun+11)=l, then for any ϵ>0 , there exists a natural number m such that | n(unun+11)l |<ϵ  nm



Therefore, we have
$$l-\epsilonNow say l+ϵ=r

Hence r<1
So we can say that $$n\left(\frac{u_n}{u_{n+1}}-1\right)

Any suggestions on how to proceed from here?



P.S. I have read other questions here on MathsSE regarding this proof specifically but I am not satisfied with these answers as one uses big O notation and another has a different form of Raabe's test. I don't want answers with big O notation. Proofs with other methods are welcome but the ones following from where I have ended are most preferred.


Answer




Your first steps are all correct.
You have obtained the inequality
unun+1<rn+1


for nm
but have realized that, in this form, it is rather hard to do anything with it.



Here is the classical argument as to how to proceed.




Rewrite this as
nunun+1(n+1)<r10.


Then observe that
nun(n+1)un+1

for all n greater than m. Thus we have an increasing
sequence {nun} and, in particular there is some positive constant c
for which nunc.




Thus for nm we have uncn


and a comparison with the harmonic series establishes
divergence as you desired.



There is a more general version of Rabbe's test known as
Kummer's test with very much the same proof.
For that you assume there is a sequence of positive numbers
Dn and you compute the limit
L=limn[Dnunun+1Dn+1]



Raabe's test is just Kummer's test with Dn=n.



The divergence part of Kummer's test (and hence also Raabe's test) doesn't actually require limits. You simply need that Dnunun+1Dn+10


for all sufficiently large n and the divergence of n=11/Dn and you can conclude divergence.


No comments:

Post a Comment

real analysis - How to find limhrightarrow0fracsin(ha)h

How to find limh0sin(ha)h without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...