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.
