# How do you use the integral test to determine if #Sigma n^ke^-n# from #[1,oo)# where k is an integer is convergent or divergent?

##### 2 Answers

The series

is convergent for any integer

#### Explanation:

Choose as test function:

This function is:

(i) non negative, as

#f(x) > 0# for#x > 0# (ii) strictly decreasing in

#(k,+oo)# as:

#f'(x) = kx^(k-1)e^(-x) -x^ke^(-x) = x^(k-1)e^(-x) (k-x) < 0# for#x > k# (iii) infinitesimal, as:

#lim_(x->oo) x^ke^(-x) = 0# (iv)

#f(n) = n^ke^(-n)#

so the convergence of the series is equivalent to the convergence of the integral:

We have to find a general formula for this integral. Start from

For

and in general:

So we can assert that:

is convergent for every

is convergent.

Application of L'Hopital's Rule to the Problem

#### Explanation:

Here is an alternative explanation:

The convergence of the "tail" (the infinitely-many terms at the end) of the series determines whether the series itself converges. We will show that the tail of the series is finite. Therefore the series converges.

Assume first that k = 0.

Then

Assume k is a POSITIVE integer.

For all x > 0,

By applying L'Hopital's rule k times, we see that ...

Therefore, for x sufficiently large,

For such values of x,

Therefore there exists a whole number, N, so that when n > N, we have

Now

Therefore, the tail of our series

Consequently,

Finally, assume k is a NEGATIVE integer.

If k < 0, then for n greater than or equal to 1,