In the ratio test, we check to see if the series #suma_k# converges or diverges by examining the ratio #a_(k+1)/a_k# as #krarroo#
If #lim_(krarroo)a_(k+1)/a_k > 1# then the series diverges.
If #0<= lim_(krarroo)a_(k+1)/a_k < 1# then the series converges.
If #lim_(krarroo)a_(k+1)/a_k = 1# then the test does not work and a new method is needed.
In this case, #a_k = ((2k)!)/k^(2k)# so
#lim_(krarroo)a_(k+1)/a_k = lim_(krarroo)(((2(k+1))!)/(k+1)^(2(k+1)))/(((2k)!)/k^(2k))#
#=>lim_(krarroo)a_(k+1)/a_k= lim_(krarroo)((2k+2)!)/((2k)!)*k^(2k)/(k+1)^(2(k+1))#
#((2k+2)!)/((2k)!) = (k+1)(k+2)# and
#k^(2k)/(k+1)^(2(k+1))=(k/(k+1))^(2k) * 1/(k+1)^2#
So, multiplying, we have
#lim_(krarroo)a_(k+1)/a_k= lim_(krarroo)(k+2)/(k+1)*(k/(k+1))^(2k)#
#lim_(krarroo)(k+2)/(k+1) = 1# and
#lim_(krarroo)(k/(k+1))^(2k) = e^(-2)# (see below for how to solve this part)
Then, multiplying gives us
#lim_(krarroo)a_(k+1)/a_k= 1*e^-2 = e^-2#
As #0 <= e^-2 < 1#, by the ratio test, the series #sum((2k)!)/k^(2k)# converges.
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To find #lim_(krarroo)(k/(k+1))^(2k) # we will first look at #lim_(krarroo)(k/(k+1))^k#
First, let's deal with the #k# exponent:
#lim_(krarroo)(k/(k+1))^k = lim_(krarroo)e^(ln((k/(k+1))^k)#
#=>lim_(krarroo)(k/(k+1))^k = lim_(krarroo)e^(kln(k/(k+1)))=e^(lim_(krarroo)kln(k/(k+1)))#
Now, looking at #lim_(krarroo)kln(k/(k+1))#, we will deal with the natural log using L'Hopital's rule.
#lim_(krarroo)kln(k/(k+1))= lim_(krarroo)ln(k/(k+1))/(1/k)#
As #lim_(krarroo)ln(k/(k+1)) = lim_(krarroo)1/k=0# we can apply L'Hopital's rule to obtain
#lim_(krarroo)ln(k/(k+1))/(1/k) = lim_(krarroo)(((k+1)/k)*(k+1-k)/(k+1)^2)/(-1/k^2)#
Simplifying, we get
#lim_(krarroo)kln(k/(k+1)) = lim_(krarroo)-k/(k+1) = -1#
Then we substitute back to obtain
#lim_(krarroo)(k/(k+1))^k = e^-1#
Thus
#lim_(karroo)(k/(k+1))^(2k) = lim_(krarroo)((k/(k+1))^k)^2 = e^-2#
