Question

Using the integral test, how do you show whether `sum (1/e^k)` diverges or converges?

Answer 1

The series converges as proved by the integral test explained below.

Explanation:

The integral test states that:

If `int_1^oo f(x)dx` converges to a value that is not infinite then `sum_(k=1)^oof(k)` will also converge.

First we have to look at the nature of `f(x) = 1/e^x`.

graph{1/e^x [-10, 10, -5, 5]}

As we can see `f(x)` is strictly decreasing from `x = 1` on words so we can apply the integral test.

Remember `f(k)= 1/e^k = e^(-k)`

Integrate this with respect to `x` to get:

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

For the upper limit, we can see that as `x` gets very large the bottom of the fraction gets also gets large, thus the fraction as a whole gets very small and vanishes completely at `x=oo` .More formally:

`lim_(x->oo)(-1/e^x)=0`

For the lower limit we simply obtain: `-1/e^`

So evaluating the limits gives:

`[-1/e^(x)]_1^oo=-1/e^` which is finite.

So, by the integral test, as the integral converges to a finite value then the summation:

`sum_(k=1)^oof(k)` also converges.

It is important to note that the integral cannot be used to evaluate the sum , but only test whether it converges or not, that is:

`sum_(k=1)^oo 1/e^k !=1/e`

Infact if we evaluate the sum we get:

`sum_(k=1)^oo 1/e^k =1/(1-e)`