Ratio Test

From ProofWiki
Jump to navigation Jump to search

Theorem

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

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

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


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

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

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

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

$\blacksquare$


Divergence

Suppose $l > 1$.

Let us take $\epsilon > 0$ small enough that $l - \epsilon > 1$.

Then, for a sufficiently large $N$, we have:

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


But $\paren {l - \epsilon}^{n - N + 1} \size {a_{N + 1} } \to \infty$ as $n \to \infty$.

So $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ diverges.

$\blacksquare$


Notes

If $l = 1$, it is impossible to say, without further analysis, whether $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ converges absolutely, converges conditionally, or diverges.

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


Sources