Question

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

Answer 1

The series converges.

Explanation:

We can apply d'Alembert's ratio test:

Suppose that;

`S=sum_(r=1)^oo a_n \ \`, and `\ \ L=lim_(n rarr oo) |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)^oo 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! ))`
`\ \ \ = lim_(n rarr oo) abs(( (-1) (n+1) )/( 2n+1 ))`
`\ \ \ = lim_(n rarr oo) abs(( n+1 )/( 2n+1 ) * (1/n)/(1/n))`
`\ \ \ = lim_(n rarr oo) abs(( 1+1/n )/( 2+1/n ))`
`\ \ \ = 1/2`

and we can conclude that the series converges