# Does the series converge or diverge? Use the integral test.

## I am confident that the given series ${\sum}_{n = 1}^{\infty} \left(\frac{1}{2} ^ n\right)$ converges, but I do not know how to prove this by using the integral test.

Mar 10, 2018

#### Explanation:

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${\sum}_{n = 1}^{\infty} \left(\frac{1}{2} ^ n\right)$

In order to prove that this series converges, using the integral test, we need to first prove that the integral of this function converges. The reason for that becomes clearer when we look at the graph of this function as shown below.

As you know, the integral of this function evaluated between specified limits gives us the area under the curve.

The difference between the Sum and the integral is that, in the case of the Sum, $x$ increases in whole integers, i.e. it indexes $1$ at a time. This means $x$ takes on values of $1 , 2 , 3 , 4 , \ldots .$ but the fractional values in between the integers are not considered in the computation, i.e. $x$ does not take on the value of $1.2 , 2.4 , \ldots$.

As a result, calculating the area under the curve, using the Sum, results in areas of a number of rectangles under the curve and adding them together, which leaves chunks of the area under the curve unaccounted for. And the number of rectangles are much fewer than in the case of the integral.

But the integral does that in a continuous manner and give us the exact area because it takes all values of $x$ into consideration and adds infinitely greater number of infinitely smaller rectangles together and produces the area which is larger than what the Sum gives us.

This is the graph of the function $y = \frac{1}{{2}^{x}}$. I have drawn three of the rectangles under the curve which represent $x = 1 , 2 , \mathmr{and} 3$. In the Sum given in this problem, it represents $n = 1 , 2 , \mathmr{and} 3$ which give us the first three values for the argument of the Sum. If we continue this process all the way to $\infty$ we will have an infinite number of rectangles whose areas will be added together to give us the value of the Sum (an inaccurate area value).

But if we integrate the function and evaluate it between $1$ and $\infty$ we will have the exact area.

If we can prove that the integral converges, i.e gives us a finite value for the area, we will prove that the Sum definitely converges because it is always smaller that what the integral gives us. Let's calculate the integral:

${\int}_{1}^{\infty} \frac{1}{{2}^{x}} \mathrm{dx}$

Let $u = - x , \therefore \mathrm{du} = - \mathrm{dx} , \therefore \mathrm{dx} = - \mathrm{du}$, substituting, we get:

$- \int {2}^{u} \mathrm{du}$

Applying the exponential rule:

$\int {a}^{u} \mathrm{du} = {a}^{u} / \ln a$, where $a = 2$ in our problem and we get:

$- \int {2}^{u} \mathrm{du} = - {2}^{u} / \ln 2$

Now, we substitute back:

int_1^oo1/2^xdx=(-2^(-x)/ln2)_1^oo=(-1/(ln2(2^x)))_1^oo=(0-(-1/(2ln2))=1/(2ln2)=0.7213475204444817

As you see, this is a finite value for the area. Therefore, the integral converges. As such, the Sum definitely converges.