Question

test for convergence or divergence n^2 * e^-n^3

Answer 1

Let's try the ratio test:

u(n+1) / u(n)

= [(n+1)^2 * e^(-(n+1)^3)] / [n^2 * e^(-n^3)]

=((n+1)/n)^2 * e^(n^3 - (n+1)^3)

 = (1 + 1/n)^2 * e^(-3n^2 - 3n - 1)

Now 1 + 1/n < e^(1/n) by considering the power series for e^(1/n)

Hence

u(n+1) / u(n)

 < e^(2/n) * e^(-3n^2 - 3n - 1)

For n>2, we know 2/n < 1

hence

u(n+1) / u(n)

< e^(1 - 3n^2 - 3n - 1)

 < e^0

Since u(n+1) / u(n) < 1

the series is convergent.

Answer 2

That response I gave is incorrect. Also it's complicated compared to the following valid solution:

xe^(-x) is convergent

therefore

(n^2) e^(-n^2) is convergent.

Now

(n^2) e^(-n^3)

 = (n^2) e^(-n^2) * [e^(n^2 - n^3)]

 < (n^2) e^(-n^2) since e^(n^2 - n^3) < 1 when n > 1

 

Therefore, by the comparison test, the series is convergent.