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Converse of the ratio test

Alexmahone

Active member
Jan 26, 2012
268
Is the converse of the ratio test true?
 

Krizalid

Active member
Feb 9, 2012
118
I don't think so. I think you can construct an easy counterexample. Care to imagine one?
 

Alexmahone

Active member
Jan 26, 2012
268
I don't think so. I think you can construct an easy counterexample. Care to imagine one?
0+0+0+... converges but the ratio is not defined.

I wonder if there are any non-trivial counterexamples.
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
0+0+0+... converges but the ratio is not defined.

I wonder if there are any non-trivial counterexamples.
The "ratio test" says that if $lim \frac{a_{n+1}}{a_n}< 1$ then $\sum a_n$ converges.

The converse is "if $\sum a_n$ converged then $lim \frac{a_{n+1}}{a_n}< 1$".

Find a convergent series such that that limit is 1.
 
Jan 31, 2012
54
0+0+0+... converges but the ratio is not defined.

I wonder if there are any non-trivial counterexamples.

Maybe...

$$ a_n=\frac{1}{n(n-1)} $$
 

Alexmahone

Active member
Jan 26, 2012
268
Find a convergent series such that that limit is 1.
$\displaystyle\sum\frac{1}{n^2}$

So, is it safe to say that if a series converges, then $\displaystyle\lim\left|\frac{a_{n+1}}{a_n}\right|\le 1$?