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

Answer 1

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.

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.

. . .

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#
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Answer 2

To use the ratio test for testing the convergence of the series ∑(2k)!/k^(2k) from n=1 to infinity, we compute the limit as n approaches infinity of the absolute value of the ratio of the (n+1)th term to the nth term. The series converges if this limit is less than 1, and diverges if it is greater than 1 or inconclusive if it equals 1.

Applying the ratio test to the given series:

We have the general term a_n = (2n)! / n^(2n)

So, the (n+1)th term is a_(n+1) = (2(n+1))! / (n+1)^(2(n+1))

Taking the ratio of the (n+1)th term to the nth term:

R = [(2(n+1))! / (n+1)^(2(n+1))] / [(2n)! / n^(2n)]

R simplifies to: R = [(2n+2)(2n+1)] / [(n+1)^2]

Taking the limit of R as n approaches infinity:

lim (n→∞) |R| = lim (n→∞) |[(2n+2)(2n+1)] / [(n+1)^2]|

= lim (n→∞) |(4n^2 + 6n + 2) / (n^2 + 2n + 1)|

= 4

Since the limit is 4, which is greater than 1, the ratio test suggests that the series diverges. Therefore, the series ∑(2k)!/k^(2k) from n=1 to infinity diverges.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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