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

Question

Elijah Abram · Accepted Answer

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

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)#

Christopher Allers · Answer

undefined To determine if the series $ \sum \frac{1}{e^k} $ converges or diverges, we use the integral test. Let $ f(x) = \frac{1}{e^x} $.

1. Check if $ f(x) $ is continuous, positive, and decreasing for all $ x \geq 1 $.
2. Integrate $ f(x) $ from 1 to infinity.
3. If the integral converges, then the series converges. If the integral diverges, then the series diverges.

1. $ f(x) = \frac{1}{e^x} $ is continuous, positive, and decreasing for $ x \geq 1 $.
2. Integrate $ f(x) $ from 1 to infinity:
$$ \int_1^\infty \frac{1}{e^x} \, dx = \lim_{b \to \infty} \int_1^b \frac{1}{e^x} \, dx = \lim_{b \to \infty} \left( -\frac{1}{e^x} \bigg|_1^b \right) = \lim_{b \to \infty} \left( -\frac{1}{e^b} + \frac{1}{e} \right) $$
$$ = \frac{1}{e} - \lim_{b \to \infty} \frac{1}{e^b} = \frac{1}{e} $$
3. Since the integral $ \int_1^\infty \frac{1}{e^x} \, dx $ converges to $ \frac{1}{e} $, the series $ \sum \frac{1}{e^k} $ also converges by the integral test.