- #1
- 2,259
- 1
It dosen't seem intuitive that if the ratio of the last two consecutive terms is less than 1 then it is convergent and viceversa if divergent when adding an infinite number of terms.
Perhaps it seems unintuitive because You think that it compares the "last" 2 terms.
I seem to think of it as since as n gets large the terms are getting smaller, ie the ratio is less than 1, than the terms must be approaching zero. Since the terms are approaching zero, it seems intuitive that it converges.
I won't prove it but I'll try to convince you that it's intuitive.
Say we have a series of general term c_n and the ratio test is conclusive. Say,
[tex]\lim_{n\rightarrow \infty} \left|\frac{c_{n+1}}{c_n}\right|=a<1[/tex]
This means that by taking n large enough, we can make |c_{n+1}| as close as we want to a*|c_n|. So, for say n>N, we can approximate |c_{N+1}| by a*|c_N|, |c_{N+2}| by a*|c_{N+1}|=a²*|c_N|, etc, such that the series looks like
[tex]\sum_{n=1}^{\infty}|c_n|=\sum_{n=1}^{N}|c_n|+|c_N|\sum_{n=1}^{\infty}a^n[/tex]
I.e. like some finite sum + a geometrical series. So the series converges absolutely.
I was thinking about proving that if the series is convergent than the ration test applies, namely a<1.
Just before doing that I think that the series in post 2 is wrong and should look like this instead:
[tex]\sum_{n=1}^{\infty}|c_n|<\sum_{n=1}^{N}|c_n|+|c_N| \sum_{n=N+1}^{\infty}a^n[/tex]