Answer:
A.
converges by integral test
Step-by-step explanation:
A. At first we need to verify that the function which the series is related (
) 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
fills these conditions:
*Considering that eˣ≠0 for all x, the function
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
is decreasing
*Because eˣ is always positive and the sum is going from 1 to ∞, this show that
is positive in [1,∞).
Now we are able to use the integral test in
as follows:
![\sum_{n=1}^{\infty}\frac{n}{e^{15n}}\ converges\ \leftrightarrow\ \int_{1}^{\infty}\frac{x}{e^{15x}}\ dx\ converges](https://tex.z-dn.net/?f=%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7Bn%7D%7Be%5E%7B15n%7D%7D%5C%20converges%5C%20%5Cleftrightarrow%5C%20%5Cint_%7B1%7D%5E%7B%5Cinfty%7D%5Cfrac%7Bx%7D%7Be%5E%7B15x%7D%7D%5C%20dx%5C%20converges)
Let´s proceed to integrate f(x) using integration by parts
![\int_{1}^{\infty}\frac{x}{e^{15x}}\ dx=\int_{1}^{\infty}xe^{-15x}\ dx](https://tex.z-dn.net/?f=%5Cint_%7B1%7D%5E%7B%5Cinfty%7D%5Cfrac%7Bx%7D%7Be%5E%7B15x%7D%7D%5C%20dx%3D%5Cint_%7B1%7D%5E%7B%5Cinfty%7Dxe%5E%7B-15x%7D%5C%20dx)
Choose your U and dV like this:
![U=x\ \rightarrow dU=1\\ dV=e^{-15x}\ \rightarrow V=\frac{-e^{-15x}}{15}](https://tex.z-dn.net/?f=U%3Dx%5C%20%5Crightarrow%20dU%3D1%5C%5C%20dV%3De%5E%7B-15x%7D%5C%20%5Crightarrow%20V%3D%5Cfrac%7B-e%5E%7B-15x%7D%7D%7B15%7D)
And continue using the formula for integration by parts:
![\int_{1}^{\infty}Udv = UV|_{1}^{\infty} - \int_{1}^{\infty}Vdu](https://tex.z-dn.net/?f=%5Cint_%7B1%7D%5E%7B%5Cinfty%7DUdv%20%3D%20UV%7C_%7B1%7D%5E%7B%5Cinfty%7D%20-%20%5Cint_%7B1%7D%5E%7B%5Cinfty%7DVdu)
![\int_{1}^{\infty}xe^{-15x}\ dx= \frac{-x}{15e^{15x}}|_{1}^{\infty} -\frac{-1}{15} \int_{1}^{\infty}e^{-15x}\ dx](https://tex.z-dn.net/?f=%5Cint_%7B1%7D%5E%7B%5Cinfty%7Dxe%5E%7B-15x%7D%5C%20dx%3D%20%5Cfrac%7B-x%7D%7B15e%5E%7B15x%7D%7D%7C_%7B1%7D%5E%7B%5Cinfty%7D%20-%5Cfrac%7B-1%7D%7B15%7D%20%5Cint_%7B1%7D%5E%7B%5Cinfty%7De%5E%7B-15x%7D%5C%20dx)
![\int_{1}^{\infty}xe^{-15x}\ dx= \frac{-x}{15e^{15x}}|_{1}^{\infty} -\frac{-1}{15}(\frac{-1}{15e^{15x}})|_{1}^{\infty}](https://tex.z-dn.net/?f=%5Cint_%7B1%7D%5E%7B%5Cinfty%7Dxe%5E%7B-15x%7D%5C%20dx%3D%20%5Cfrac%7B-x%7D%7B15e%5E%7B15x%7D%7D%7C_%7B1%7D%5E%7B%5Cinfty%7D%20-%5Cfrac%7B-1%7D%7B15%7D%28%5Cfrac%7B-1%7D%7B15e%5E%7B15x%7D%7D%29%7C_%7B1%7D%5E%7B%5Cinfty%7D)
![\int_{1}^{\infty}xe^{-15x}\ dx= \frac{-x}{15e^{15x}}|_{1}^{\infty} -\frac{1}{225e^{15x}}|_{1}^{\infty}](https://tex.z-dn.net/?f=%5Cint_%7B1%7D%5E%7B%5Cinfty%7Dxe%5E%7B-15x%7D%5C%20dx%3D%20%5Cfrac%7B-x%7D%7B15e%5E%7B15x%7D%7D%7C_%7B1%7D%5E%7B%5Cinfty%7D%20-%5Cfrac%7B1%7D%7B225e%5E%7B15x%7D%7D%7C_%7B1%7D%5E%7B%5Cinfty%7D)
Because we are dealing with ∞, we´d rewrite it as a limit that will help us at the end of the integral:
![\int_{1}^{\infty}xe^{-15x}\ dx= \lim_{b \to{\infty}}(\frac{-x}{15e^{15x}}|_{1}^{b}-\frac{1}{225e^{15x}}|_{1}^{b})](https://tex.z-dn.net/?f=%5Cint_%7B1%7D%5E%7B%5Cinfty%7Dxe%5E%7B-15x%7D%5C%20dx%3D%20%5Clim_%7Bb%20%5Cto%7B%5Cinfty%7D%7D%28%5Cfrac%7B-x%7D%7B15e%5E%7B15x%7D%7D%7C_%7B1%7D%5E%7Bb%7D-%5Cfrac%7B1%7D%7B225e%5E%7B15x%7D%7D%7C_%7B1%7D%5E%7Bb%7D%29)
![\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}})](https://tex.z-dn.net/?f=%5Cint_%7B1%7D%5E%7B%5Cinfty%7Dxe%5E%7B-15x%7D%5C%20dx%3D%20%5Clim_%7Bb%20%5Cto%7B%5Cinfty%7D%7D%20%5Cfrac%7B-b%7D%7B15e%5E%7B15b%7D%7D-%5Cfrac%7B1%7D%7B225e%5E%7B15b%7D%7D-%28%5Cfrac%7B-1%7D%7B15e%5E%7B15%7D%7D-%5Cfrac%7B1%7D%7B225e%5E%7B15%7D%7D%29)
![\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})](https://tex.z-dn.net/?f=%5Cint_%7B1%7D%5E%7B%5Cinfty%7Dxe%5E%7B-15x%7D%5C%20dx%3D%20%28%20%5Clim_%7Bb%20%5Cto%7B%5Cinfty%7D%7D%20%5Cfrac%7B-b%7D%7B15e%5E%7B15b%7D%7D-%5Cfrac%7B1%7D%7B225e%5E%7B15b%7D%7D%29%2B%5Cfrac%7B1%7D%7B15e%5E%7B15%7D%7D%281-%5Cfrac%7B1%7D%7B15%7D%29)
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
![\lim_{b \to{\infty}} \frac{-b}{15e^{15b}}-\frac{1}{225e^{15b}} = 0](https://tex.z-dn.net/?f=%5Clim_%7Bb%20%5Cto%7B%5Cinfty%7D%7D%20%5Cfrac%7B-b%7D%7B15e%5E%7B15b%7D%7D-%5Cfrac%7B1%7D%7B225e%5E%7B15b%7D%7D%20%3D%200)
![\int_{1}^{\infty}xe^{-15x}\ dx= \frac{1}{15e^{15}}(1-\frac{1}{15})](https://tex.z-dn.net/?f=%5Cint_%7B1%7D%5E%7B%5Cinfty%7Dxe%5E%7B-15x%7D%5C%20dx%3D%20%5Cfrac%7B1%7D%7B15e%5E%7B15%7D%7D%281-%5Cfrac%7B1%7D%7B15%7D%29)
Showing that the integral converges, it´s the same as showing that the series converges.
By the integral test
converges