Finding convergence of this series using Integral/Comparison

Homework Statement

series from n = 1 to infinity, (ne^(-n))

The Attempt at a Solution

I want to use integral test.
I know this function is:
positive (on interval 1 to infinity)
continous
and finding derivative of f(x) = xe^(-x) I found it to be ultimately decreasing.

So integral test is applicable.

I set up integral from 1 to infinity (xe^(-x))

u = x du = dx
v = -e^(-x) dv = e^(-x)

-xe^(-x) + integral e^(-x)

-xe^(-x) - e^(-x) = -e^(-x) (x + 1)

evaluating from 1 to t

$-(1/e^t) (t+1) + 2/e^(1)$

but now when I do lim t -> infinity, -(1/e^t) (t+1) should = infinity/infinity, which would mean $a_n$ would be divergent, but it is convergent.

does anyone know where my mistake is??

Mark44
Mentor

Homework Statement

series from n = 1 to infinity, (ne^(-n))

The Attempt at a Solution

I want to use integral test.
I know this function is:
positive (on interval 1 to infinity)
continous
and finding derivative of f(x) = xe^(-x) I found it to be ultimately decreasing.

So integral test is applicable.

I set up integral from 1 to infinity (xe^(-x))

u = x du = dx
v = -e^(-x) dv = e^(-x)

-xe^(-x) + integral e^(-x)

-xe^(-x) - e^(-x) = -e^(-x) (x + 1)

evaluating from 1 to t

$-(1/e^t) (t+1) + 2/e^(1)$
I get something slightly different from what you got, namely ##te^{-t} - e^{-t} + 2e^{-1} = (t - 1)e^{-t} + 2e^{-1}##
, which would mean $a_n$ would be divergent, but it is convergent.