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