How do you use the ratio test to test the convergence of the series #∑(2k)!/k^(2k) # from n=1 to infinity?

1 Answer
Nov 7, 2015

Answer:

Take the limit as #krarroo# of a general term in the series divided by its preceding term, and apply the rules of the ratio test to see that the series converges.

Explanation:

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#