How do you test the series #Sigma (2^n(n!))/n^n# from #n=[1,oo)# by the ratio test?

1 Answer
Jan 21, 2018

The series converges.

See below.

Explanation:

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.