ratio test of d’Alembert

A lighter version of the ratio test is the

Ratio test of d’Alembert.  Let  $a_1+a_2+\mathrm{\dots }$  be a series with positive terms.

$1^{\circ }$.  If there exists a number $q$ such that    and

${\displaystyle \frac{a_{n+1}}{a_n}}\le q\mathit{\hspace{1em}}\text{for all}n\ge n_0,$(1)

then the series converges.

$2^{\circ }$.  If there exists a number $n_0$ such that

${\displaystyle \frac{a_{n+1}}{a_n}}\ge \mathrm{\hspace{0.33em}1}\mathit{\hspace{1em}}\text{for all}n\ge n_0,$(2)

then the series diverges.

Proof.  $1^{\circ }$. By the condition (1), we have  $a_{n+1}\le a_{n}q$;  thus we get the estimations

$$a_{n_0+1}\le a_{n_0}q,$$

$$a_{n_0+2}\le a_{n_0+1}q\le a_{n_0}{q}^2,$$

$$\mathrm{\cdots }\mathit{\hspace{1em}}\mathrm{\cdots }\mathit{\hspace{1em}}\mathrm{\cdots }$$

$$a_{n_0+p}\le a_{n_0+p-1}q\le \mathrm{\dots }\le a_{n_0}{q}^p,$$

$$\mathrm{\cdots }\mathit{\hspace{1em}}\mathrm{\cdots }\mathit{\hspace{1em}}\mathrm{\cdots }$$

Because  $a_{n_0}q+a_{n_0}{q}^2+\mathrm{\dots }+a_{n_0}{q}^p+\mathrm{\dots }$  is a convergent geometric series, those inequalities and the comparison test imply that the series

$$a_{n_0+1}+a_{n_0+2}+\mathrm{\dots }+a_{n_0+p}+\mathrm{\dots }$$

and as well the whole series  $a_1+a_2+\mathrm{\dots }$  is convergent.

$2^{\circ }$.  The condition (2) yields

$$a_{n_0+1}\ge a_{n_0},a_{n_0+2}\ge a_{n_0+1}\ge a_{n_0},\mathrm{\dots }$$

and since  $a_{n_0}$ is positive, the limit of $a_n$ as $n$ tends to infinity cannot be 0.  Hence the given series does not fulfil the necessary condition of convergence.

Example.  If the variable $x$ in the power series

$$\sum _{n=0}^{\mathrm{\infty }}n!x^n$$

is distinct from zero, we have

$$\frac{|(n+1)!x^{n+1}|}{|n!x^n|}=(n+1)|x|\ge \mathrm{\hspace{0.33em}1}\mathit{\hspace{1em}}\text{for all}n\ge n_0.$$

Then the series does not converge absolutely (http://planetmath.org/AbsoluteConvergence).  The known theorem of Abel says that the series diverges for all $x\ne 0$.  It means that the radius of convergence is 0.