Math Questions and Answers

Start Your Free Trial

`sum_(n=1)^oo n^k*e^(-n)` Use the Integral Test to determine the convergence or divergence of the series, where k is a positive integer.

Expert Answers info

pnrjulius eNotes educator | Certified Educator

calendarEducator since 2016

write550 answers

starTop subjects are History, Science, and Business

The Integral test says that this sum will converge if and only if this integral also converges:`int_{1}^{infty} x^k e^{-x} dx `

When integrating this, we would use integration by parts, and we would need to use it k times. The first set of parts is
`u = x^k, dv = e^{-x} dx, du = k x^{k-1} dx, v = -e^{-x}`

 

`int u dv = u v - int v du = - x^k e^{-x}|_1^infty + int_{1}^{infty} k x^{k-1} e^{-x} dx `

Then we repeat for `u_1 = x^{k-1}` , and so on until we have only the `e^{-x}` term left.


But the important thing is that the last term would only be in terms of a constant times `int e^-x dx` , which clearly converges; and then all the other terms would look like this, for some integer `1 leq p leq k`
and some constant C:


`C x^p e^{-x} |_{1}^{infty}`

The value of this term at `x = 1 ` we can simply calculate; no problem there, it will be some finite number. The limit as x goes to infinity we can also determine by the fact that `e^x` always increases faster than any polynomial as x gets very large, and thus for any value of p, this limit must be zero.

Thus, we have k-1 terms that are finite (zero minus a finite value), plus one final term that is a convergent integral. Therefore the whole integral converges; therefore the sum converges.

check Approved by eNotes Editorial