Ratio Test

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:

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.

Divergence
Suppose $l > 1$.

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

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


 * $\displaystyle \size{\size{\frac{a_{k + 1}}{a_k}} - l} < \epsilon \quad\leadsto\quad \size{\frac{a_{k + 1}}{a_k}} > l - \epsilon$

and hence, for all $n \ge N$:

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

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