Ratio Test

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Theorem

Let $\ds \sum_{n \mathop = 1}^\infty a_n$ be a series of real numbers in $\R$, or a series of complex numbers in $\C$.

Let the sequence $\sequence {a_n}$ satisfy:

$\ds \lim_{n \mathop \to \infty} \size {\frac {a_{n + 1} } {a_n} } = l$


If $l > 1 $, then $\ds \sum_{n \mathop = 1}^\infty a_n$ diverges.
If $l < 1 $, then $\ds \sum_{n \mathop = 1}^\infty a_n$ converges absolutely.


Proof

From the statement of the theorem, it is necessary that $\forall n: a_n \ne 0$; otherwise $\size {\dfrac {a_{n + 1} } {a_n} }$ is not defined.

Here, $\size {\dfrac {a_{n + 1} } {a_n} }$ denotes either the absolute value of $\dfrac {a_{n + 1} } {a_n}$, or the complex modulus of $\dfrac {a_{n + 1} } {a_n}$.


Absolute Convergence

Suppose $l < 1$.

Let us take $\epsilon > 0$ such that $l + \epsilon < 1$.

Then:

$\exists N: \forall n > N: \size {\dfrac {a_n} {a_{n - 1} } } < l + \epsilon$

Thus:

\(\ds \size {a_n}\) \(=\) \(\ds \size {\frac {a_n} {a_{n - 1} } } \size {\frac {a_{n - 1} } {a_{n - 2} } } \dotsm \size {\frac {a_{N + 2} } {a_{N + 1} } } \size {a_{N + 1} }\)
\(\ds \) \(<\) \(\ds \paren {l + \epsilon}^{n - N - 1} \size {a_{N + 1} }\)

By Sum of Infinite Geometric Progression, $\ds \sum_{n \mathop = 1}^\infty \paren {l + \epsilon}^n$ converges.

So by the the corollary to the comparison test, it follows that $\ds \sum_{n \mathop = 1}^\infty \size {a_n}$ converges absolutely too.

$\blacksquare$


Divergence

Suppose $l > 1$.

Let $\epsilon > 0$ such that $l - \epsilon > 1$.

Then, for sufficiently large $N$, for all $k \ge N$ we have:

\(\ds \size {\size {\frac {a_{k + 1} } {a_k} } - l}\) \(<\) \(\ds \epsilon\)
\(\ds \leadsto \ \ \) \(\ds \size {\frac {a_{k + 1} } {a_k} }\) \(>\) \(\ds l - \epsilon\)


Hence, for all $n \ge N$:

\(\ds \size {a_n}\) \(=\) \(\ds \size {\frac {a_n} {a_{n - 1} } } \size {\frac {a_{n - 1} } {a_{n - 2} } } \dotsm \size {\frac {a_{N + 1} } {a_{N} } } \size {a_{N} }\)
\(\ds \) \(>\) \(\ds \paren {l - \epsilon}^{n - N} \size {a_{N} }\)


But:

$\paren {l - \epsilon}^{n - N} \size {a_{N} } \to \infty$

as $n \to \infty$.

So, by the Divergence Test, $\ds \sum_{n \mathop = 1}^\infty a_n$ diverges.

$\blacksquare$


Warning

If $l = 1$, the Ratio Test provides no information on whether $\ds \sum_{n \mathop = 1}^\infty a_n$ converges absolutely, converges conditionally, or diverges.

If $\size {\dfrac {a_{n + 1} } {a_n} } \to \infty$ as $n \to \infty$, then of course $\ds \sum_{n \mathop = 1}^\infty a_n$ diverges.


Also known as

The Ratio Test is also seen referred to as the Cauchy Ratio Test, after Augustin Louis Cauchy.

It can also be seen as D'Alembert's Test, after Jean le Rond d'Alembert.


Sources

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