Finding convergence of this series using Integral/Comparison

[Rijad Hadzic]

Homework Statement

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

Homework Equations

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]

Homework Statement

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

Homework Equations

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 $te^{-t} - e^{-t} + 2e^{-1} = (t - 1)e^{-t} + 2e^{-1}$

but now when I do lim t -> infinity, -(1/e^t) (t+1) should = infinity/infinity

But the fact that you're getting the indeterminate form $[\frac \infty \infty]$ doesn't tell you anything. You can use L'Hopital's Rule to actually evaluate your limit, which results in an actual value for this limit.

, which would mean a_n would be divergent, but it is convergent.

does anyone know where my mistake is??

 

[Mark44]

BTW, this thread was marked as "Solved" but that didn't appear to really be the case, so I have changed it to "Unsolved."