Question

How do you use the integral test to determine if `Sigma e^-n` from `[1,oo)` is convergent or divergent?

Answer 1

The series:

`sum_(n=1)^oo e^(-n)`

is convergent.

Explanation:

We can apply the integral test using as test function:

`f(x) = e^(-x)`

The function is positive, strictly decreasing, infinitesimal for `x->oo` and we have `f(n) = e^(-n)`, so the convergence of the series is equivalent to the convergence of the indefinite integral:

`int_1^oo e^(-x)dx = [-e^(-x)]_1^oo = 1/e`

and the series is proven to be convergent.