Expert: Sherry Wallin - 3/21/2011

Question
QUESTION: Hello Dr. Wallin how are you?

I have to find if the following series converge:

a) Sum from k=1 to k=inf of k^k*e^(-k^2)

b) Sum from k=1 to k=inf of (1/sqrt(k))*e^(-2*sqrt(k))

For a, I told that since k^k*e^(-k^2) is not equal to zero for k>=1 we can use the ratio test. I let a_k=k^k*e^(-k^2) then a_(k+1)/a_k=(((k+1)/k)^k)*(k+1) * e^(-2k+1) after some calculations. Then lim as k tends to inf of a_(k+1)/a_k is equal to 0<1 so our series converges absolutely by the ratio test.

c) I also have to determine the radius of convergence of the power series G(x)=Sum from k=1 to inf of (2k choose k)*x^k

For this I used the ratio test and found that abs(a_(k+1)/a_k)= 4 * abs(x) * (k+1/2)/(k+1) and as k tends to infinity then abs(a_(k+1)/a_k) tends to 4*abs(x)
Thus G(x) diverges for abs(x)>1/4 and converges absolutely for abs(x) < 1/4 so radius of convergence = 1/4

So basically if you can check part a and c if I am correct and to guide me and give me the solution for part b.

Thanks in advance for any help!

ANSWER: Michael~


You have a_(k+1)/a_k=(((k+1)/k)^k)*(k+1) * e^(-2k+1)  and it should be
a_(k+1)/a_k=(((k+1)/k)^k)*(k+1) * e^(-2k-1)
I agree that lim as k tends to infinity of a_(k+1)/a_k is 0.


For b:  use the ratio test also
(1/sqrt(k+1))*e^(-2*sqrt(k+1))/(1/sqrt(k))*e^(-2*sqrt(k)) = 0 as k tends to infinity.

I don't have time to check part c right now, sorry..., I can do that later if you still want me to

Math Prof



---------- FOLLOW-UP ----------

QUESTION: Hello Dr. Wallin again, I tried to do part b using the ratio test but found that abs(a_(k+1)/a_k) = abs(sqrt(k/(k+1)) * e^ (2*(sqrt(k) * sqrt(k+1))
which tends to 1 as k tends to infinity, so I can't deduce if the sequence converges or diverges using the ratio test.
Hoe did you find it to be equal with zero?

Answer
Michael~

My bad, you are right the series does tend to 1 and not 0 so my recommendation is for you to use the integral test.
Let f(x) = 1/(sqrt x)*e^(-2*sqrt x)= 1/(x^(1/2))e^(-2x^(1/2))= 1/(x^.5)[e^(-2x^.5)]

Int(1,00,x,dx) means the integral over the limits 1 to infinity, for the function f, with respect to x
Int(1,00,f,dx) (1/(x^.5)[e^(-2x^.5)])dx = lim (t->00)Int(1,t,f,t)(1/(x^.5)[e^(-2x^.5)])
let u = [-2x^.5]then du = -x^(-.5) dx => - du = x^(-.5) = 1/(x^.5) thus you are integrating and taking the limit of
lim(t->00)Int(1,t)[-e^u du]
= lim(t->00)-e^(-2x^.5) => -e^(-2t^.5)-(-e^-2(00)^.5)
= -1/(e^2t^.5)+ 1/e^(2*00)^.5
= -1/(e^2t^.5)+ 0  since e to a huge number is huge and it's reciprocal will go to 0
= -1/(e^2t^.5) is an actual value and means that integral converges
Therefore by the integral test the series 1/(sqrt k)*e^(-2*sqrt k) converges also.

Math Prof