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 $\left \langle {a_n} \right \rangle$ satisfy $\displaystyle \lim_{n \to \infty} \left\vert{\frac {a_{n+1}}{a_n}}\right\vert = l$.


 * 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 $\left\vert{\dfrac {a_{n+1}}{a_n}}\right\vert$ is not defined.

Here, $\left \vert {\dfrac {a_{n+1}}{a_n}} \right \vert$ 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:
 * $\exists N: \forall n > N: \left\vert{\dfrac {a_n}{a_{n-1}}}\right\vert < l + \epsilon$

Thus:

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

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

Divergence
Suppose $l > 1$.

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

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

But $\left({l - \epsilon}\right)^{n - N + 1} \left\vert{a_{N+1}}\right\vert \to \infty$ as $n \to \infty$.

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