# How do you determine the convergence or divergence of Sigma ((-1)^n n!)/(1*3*5***(2n-1) from [1,oo)?

Sep 19, 2017

The series converges.

#### Explanation:

We can apply d'Alembert's ratio test:

Suppose that;

$S = {\sum}_{r = 1}^{\infty} {a}_{n} \setminus \setminus$, and $\setminus \setminus L = {\lim}_{n \rightarrow \infty} | {a}_{n + 1} / {a}_{n} |$

Then

if L < 1 then the series converges absolutely;
if L > 1 then the series is divergent;
if L = 1 or the limit fails to exist the test is inconclusive.

So our series is;

$S = {\sum}_{n = 0}^{\infty} {a}_{n}$ where a_n = ( (-1)^n n! )/( 1.3.5 ... (2n-1) )

So our test limit is:

 L = lim_(n rarr oo) abs(( ( (-1)^(n+1) (n+1)! )/( 1.3.5 ... (2n-1).(2(n+1)-1) ) ) / ( ( (-1)^n n! )/( 1.3.5 ... (2n-1) ) ))
 \ \ \ = lim_(n rarr oo) abs(( (-1)^(n+1) (n+1)! )/( 1.3.5 ... (2n-1).(2(n+1)-1) ) * ( 1.3.5 ... (2n-1) ) / ( (-1)^n n! ) )
 \ \ \ = lim_(n rarr oo) abs(( (-1) n!(n+1) )/( 2n+2-1 ) * ( 1 ) / ( n! ))
$\setminus \setminus \setminus = {\lim}_{n \rightarrow \infty} \left\mid \frac{\left(- 1\right) \left(n + 1\right)}{2 n + 1} \right\mid$
$\setminus \setminus \setminus = {\lim}_{n \rightarrow \infty} \left\mid \frac{n + 1}{2 n + 1} \cdot \frac{\frac{1}{n}}{\frac{1}{n}} \right\mid$
$\setminus \setminus \setminus = {\lim}_{n \rightarrow \infty} \left\mid \frac{1 + \frac{1}{n}}{2 + \frac{1}{n}} \right\mid$
$\setminus \setminus \setminus = \frac{1}{2}$

and we can conclude that the series converges