This is an alternating series in the form:
#sum_n (-1)^n a_n#
with:
#a_n = (n!2^n)/(2n!) >0#
Write the generic element of the sequence #{a_n}# as:
#a_n = (n!2^n)/(2n!) = 2^n/(2n(2n-1)...(n+1))#
#a_n = underbrace( (2/(2n)) (2/(2n-1)) ... (2/(n+1)))_"n factors"#
#a_n = underbrace( (1/n) (1/(n-1/2)) ... (1/(n-(n-1)/2)))_"n factors"#
#a_n = prod_(k=0)^(n-1) 1/(n-k/2)#
We can then see that:
#(1) " " lim_(n->oo) a_n = 0#
and that, as:
#a_(n+1) = prod_(k=0)^n 1/(n+1-k/2)#
if we look at the products defining #a_n# and #a_(n+1)#, we see that for the same value of #k#:
#1/(n+1-k/2) < 1/(n-k/2)#
and in #a_(n+1)# we have an additional factor:
#1/(n+1-n/2) = 1/(n/2+1) = 2/(n+2) <= 1#
We can conclude that:
#(2) " " a_(n+1) < a_n#
and then based on Leibniz' theorem the conditions #(1)# and #(2)# are sufficient to determine that the series is convergent.
