We have:
#sum_(n=1)^oo(2^n(n!))/(n^n)#
The ratio test tells us the series will converge if :
#lim_(n->oo)abs(a_(n+1)/a_n)<1#
In this case: #a_n = (2^n(n!))/(n^n)#
So:
#lim_(n->oo)abs(a_(n+1)/a_n)=lim_(n->oo)abs((2^(n+1)(n+1)!)/((n+1)^(n+1)))/abs((2^(n)(n)!)/(n^n))#
#=lim_(n->oo)abs((n^n2^(n+1)(n+1)!)/((n+1)^(n+1)2^n n!))#
We can do a bit of cancelling here:
Note that: #2^(n+1)/2^n=2^(n+1-n)=2#
And #((n+1)!)/(n!)=((n+1)timesntimes...times3times2times1)/(ntimes...times3times2times1)#
#=((n+1)timescancel(ntimes...times3times2times1))/cancel(ntimes...times3times2times1)=n+1#
So we can simplify the limit a bit:
#=lim_(n->oo)abs((n^ncolor(blue)(2^(n+1))color(green)((n+1)!))/((n+1)^(n+1)color(blue)(2^n) color(green)(n!)))#
#=lim_(n->oo)abs((n^ncolor(blue)2color(green)((n+1)))/((n+1)^(n+1)))=lim_(n->oo)abs(2(n^n(n+1))/((n+1)^(n+1)))#
We can now cancel the #n+1# on the top with the power of #n+1# on the bottom like so:
#lim_(n->oo)abs(2(n^ncolor(red)((n+1)))/((n+1)^(n+color(red)1)))=lim_(n->oo)abs(2(n^n)/((n+1)^n))#
Obviously for integers #n# and #n>0# this will always be real and positive so there is no need for the "absolute" brackets;
#=2lim_(n->oo)(n^n)/((n+1)^n)#
To evaluate this limit consider:
#L = lim_(n->oo)(n^n)/((n+1)^n)#
So:
#ln(L) = ln(lim_(n->oo)(n^n)/((n+1)^n))=lim_(n->oo)ln(((n)/(n+1))^n)#
#lim_(n->oo)n ln((n)/(n+1))=lim_(n->oo)-nln((n+1)/n)#
#=lim_(n->oo)-nln(1+1/n)=-lim_(n->oo)ln(1+1/n)/((1/n))->0/0#
So use L'Hopital's rule:
#d/(dn)ln(1+1/n)=1/(1+1/n)*(-1/n^2)#
#d/(dn)(1/n)=-1/n^2#
So limit now becomes:
#-lim_(n->oo)ln(1+1/n)/((1/n))=-lim_(n->oo)(1/(1+1/n)*(-1/n^2))/(-1/n^2)#
The factor of #(-1/n^2)# cancels to give:
#-lim_(n->oo)1/(1+1/n)=-1/(1+0)=-1#
So #ln(L)=-1# so it follows that:
#L = lim_(n->oo)(n^n)/((n+1)^n) = e^(-1)=1/e#
Hence:
#2lim_(n->oo)(n^n)/((n+1)^n)=2/e#
Finally, it follows that:
#lim_(n->oo)abs(a_(n+1)/a_n)=lim_(n->oo)abs((n^n2^(n+1)(n+1)!)/((n+1)^(n+1)2^n n!))=2/e<1#
So by the ratio test the series converges.
