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Proof of the ratio test 
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#1
Nov712, 12:33 AM

P: 783

I am trying to understand something in the proof of the ratio test for series convergence.
If [itex]a_{n}[/itex] is a sequence of positive numbers, and that the ratio test shows that [itex] \lim_{n→∞}\frac{a_{n+1}}{a_{n}} = r < 1[/itex], then the series converges. Apparently, the proof defines a number R : r<R<1, and then shows that there exists a N>0 such that [itex]\frac{a_{n+1}}{a_{n}} < R [/itex] for all n>N. It need not to be true in the case where n=N, right? Up to this part I get. But then it concludes from the above that, there exists a positive N such that [itex] a_{N+1}<a_{n}R [/itex] which does not follow due to the statement in bold. Could someone please point out where I am wrong so I can continue this theorem without any qualms? Thanks! BiP 


#2
Nov712, 01:05 AM

P: 439

Hmm, think about what it means for the limit to exist. I think that might be the missing piece that shows why it works. Because if you believe it exist, there is an n greater than or equal to N such that (a_(n+1))/(a_n) is less than or equal to R for n greater than N. Then from there you can rewrite the inequality to get what you got.



#3
Nov712, 02:36 PM

P: 783

BiP 


#4
Nov712, 02:40 PM

Mentor
P: 18,099

Proof of the ratio test
For all [itex]\varepsilon>0[/itex], there exists an N such that for all [itex]n\geq N[/itex] holds that [itex]a_na<\varepsilon[/itex]. is actually equivalent with For all [itex]\varepsilon>0[/itex], there exists an N such that for all [itex]n> N[/itex] holds that [itex]a_na<\varepsilon[/itex]. So you can use both statements to define limit of a sequence. Of course, once you decided on which of both versions to use, you have to be consistent. 


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