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I know the D'Alembert criterion for series, but I was wondering if it applied for sequences too. This is, if $a_n$ is a sequence, then let

$$\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=L$$

If $L<1$ the sequence converges; if $L=1$ we can't say anything about its convergence; if $L>1$ the sequence diverges.

Is this true?

Bernard
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lafinur
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  • There is a "ratio test for sequences" (though the theorem usually doesn't have a name). See here: https://math.stackexchange.com/questions/2516157/why-does-the-ratio-test-prove-that-this-particular-sequence-converges-on-0 – bjorn93 Jan 13 '20 at 19:28
  • As stated, the criterion is not true (see my counter example below). You need to incorporate absolute values in the test, or assume - e. g. - that you are only working with positive real numbers. The link in the comment of @bjorn93 is very useful for that matter. – Boccherini Mar 31 '20 at 14:04

3 Answers3

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Yes, It's true and can be proved like you prove the criteria for series (by taking geometrical sequence based on L and compare it to $a_n$). You can also shorten it by using the criteria for series, if $\lim_{n \to \inf} \frac{a_{n+1}}{a_n} = L $ (< 1 for example) then by D'Alembert criteria $\sum_{n = 1}^{\inf}a_n$ converges and therefore $a_n \to 0$

Daniel Segal
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If $L<1$ D'Alembert for series says that $\sum a_n$ is convergent, hence $a_n\to0$.

If $L=1$, take $a_n=n$ (divergent) $a_n=1/n$ convergent. So you cannot conclud anything.

If $L>1$, take $K\in (1,L)$. You know that there is $N\in\Bbb N$ such that $\frac{a_{n+1}}{a_n}>K>1$, so $a_{n+1}>K a_n>K^2a_{n-1}>\ldots > K^{n-N+1} a_N$. But for $K>1$, $K^{n+1-N}\to+\infty$. So, because $a_n>K^{n-N+1}a_N$, you have that $a_n\to\infty$.

Obviously, all this stuff if $a_n\ge0$ for all $n\in\Bbb N$.

Tito Eliatron
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No, it is not true.

Look for instance at the sequence $$1, -2, 4, -8, 16, \ldots$$

We have that $$L = \lim_{n \mapsto +\infty}\frac{a_{n + 1}}{a_n} = -2 < 1$$ and the sequence does not converge, nor diverge.

Boccherini
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