MHB Does the Converse of the Ratio Test Always Hold for Convergent Series?

alexmahone
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Is the converse of the ratio test true?
 
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I don't think so. I think you can construct an easy counterexample. Care to imagine one?
 
Krizalid said:
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.
 
Alexmahone said:
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.
 
Alexmahone said:
0+0+0+... converges but the ratio is not defined.

I wonder if there are any non-trivial counterexamples.
Maybe...

$$ a_n=\frac{1}{n(n-1)} $$
 
HallsofIvy said:
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 $\displaystyle\lim\left|\frac{a_{n+1}}{a_n}\right|\le 1$?
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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