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

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)#