# Root test inconclusive implies ratio test inconclusive

I’m a high school student and teaching myself a tiny bit of maths. I stumbled over a problem, I can’t find a proof for (and neither found a proof for it in the internet or in books, but I guess there is one).

If the root test is inconclusive, then the ratio test is inconclusive, too ( in other words: if the root test equals one, then the ratio test equals one).

Is this statement true? And is there a proof for it?

Any help is appreciated.

(The point of the proof should be showing that the root test is stronger than the ratio test. I know examples for this, but can't find a proof (maybe proof by contradiction?).)

What I want to show is:

$$lim sup \sqrt[n]{|a_n|}=1 \implies lim |\frac{a_{n+1}}{a_n}|=1$$

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• Jun 23 at 7:50
• Only partly (or maybe I just don’t see it). The conclusion I can make from this, is just that if I the root test equals 1, then the ratio test could equal one (or be greater). Jun 23 at 10:24
• Jun 23 at 13:40
• But how does this prove my statement $lim sup \sqrt[n]{|a_n|}=1 \implies lim |\frac{a_{n+1}}{a_n}|=1$ ? Jun 23 at 14:16