Use the Integral Test to determine whether the series is convergent or divergent
[infinity]
Σne−15n
n = 1
In order to apply the Integral Test for the function
f(x)
on the interval [a, [infinity]), it must be true that (choose all that apply)
a. f must be continuous.
b. f must be differentiable.
c. f must be positive.
d. f must be negative.
e. f must be increasing.
f. f must be decreasing.

A. At first we need to verify that the function which the series is related ([tex]\frac{n}{e^{15n}}[/tex]) fills the necessary conditions to ensure that the test is effective.

*f(x) must be continuous or differentiable

*f(x) must be positive and decreasing

Let´s verify that [tex]f(x)=\frac{n}{e^{15n}}[/tex] fills these conditions:

*Considering that eˣ≠0 for all x, the function [tex]f(x)=\frac{n}{e^{15n}}[/tex] does not have any discontinuities, so it´s continuous

*Because eˣ is increasing:

if a<b ,then eᵃ<eᵇ

if 0<eᵃ<eᵇ ,then 1/eᵃ > 1/eᵇ

if 1/eᵃ > 1/eᵇ and a<b, then a/eᵃ<b/eᵇ

We conclude that [tex]f(x)=\frac{n}{e^{15n}}[/tex] is decreasing

*Because eˣ is always positive and the sum is going from 1 to ∞, this show that [tex]f(x)=\frac{n}{e^{15n}}[/tex] is positive in [1,∞).

Now we are able to use the integral test in [tex]f(x)=\frac{n}{e^{15n}}[/tex] as follows:

[tex]\sum_{n=1}^{\infty}\frac{n}{e^{15n}}\ converges\ \leftrightarrow\ \int_{1}^{\infty}\frac{x}{e^{15x}}\ dx\ converges[/tex]

Let´s proceed to integrate f(x) using integration by parts

[tex]\int_{1}^{\infty}\frac{x}{e^{15x}}\ dx=\int_{1}^{\infty}xe^{-15x}\ dx[/tex]

Choose your U and dV like this:

[tex]U=x\ \rightarrow dU=1\\ dV=e^{-15x}\ \rightarrow V=\frac{-e^{-15x}}{15}[/tex]

And continue using the formula for integration by parts:

[tex]\int_{1}^{\infty}Udv = UV|_{1}^{\infty} - \int_{1}^{\infty}Vdu[/tex]

[tex]\int_{1}^{\infty}xe^{-15x}\ dx= \frac{-x}{15e^{15x}}|_{1}^{\infty} -\frac{-1}{15} \int_{1}^{\infty}e^{-15x}\ dx[/tex]

[tex]\int_{1}^{\infty}xe^{-15x}\ dx= \frac{-x}{15e^{15x}}|_{1}^{\infty} -\frac{-1}{15}(\frac{-1}{15e^{15x}})|_{1}^{\infty}[/tex]

[tex]\int_{1}^{\infty}xe^{-15x}\ dx= \frac{-x}{15e^{15x}}|_{1}^{\infty} -\frac{1}{225e^{15x}}|_{1}^{\infty}[/tex]

Because we are dealing with ∞, we´d rewrite it as a limit that will help us at the end of the integral:

[tex]\int_{1}^{\infty}xe^{-15x}\ dx= \lim_{b \to{\infty}}(\frac{-x}{15e^{15x}}|_{1}^{b}-\frac{1}{225e^{15x}}|_{1}^{b})[/tex]

[tex]\int_{1}^{\infty}xe^{-15x}\ dx= \lim_{b \to{\infty}} \frac{-b}{15e^{15b}}-\frac{1}{225e^{15b}}-(\frac{-1}{15e^{15}}-\frac{1}{225e^{15}})[/tex]

[tex]\int_{1}^{\infty}xe^{-15x}\ dx= ( \lim_{b \to{\infty}} \frac{-b}{15e^{15b}}-\frac{1}{225e^{15b}})+\frac{1}{15e^{15}}(1-\frac{1}{15})[/tex]

We only have left to solve the limits, but because b goes to ∞ and it is in an exponential function on the denominator everything goes to 0

[tex]\lim_{b \to{\infty}} \frac{-b}{15e^{15b}}-\frac{1}{225e^{15b}} = 0[/tex]

[tex]\int_{1}^{\infty}xe^{-15x}\ dx= \frac{1}{15e^{15}}(1-\frac{1}{15})[/tex]

Showing that the integral converges, it´s the same as showing that the series converges.

By the integral test [tex]\sum_{n=1}^{\infty}\frac{n}{e^{15n}}[/tex] converges