`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.

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...

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.