Finding convergence of this series using Integral/Comparison

  • #1
321
20

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

[itex] -(1/e^t) (t+1) + 2/e^(1) [/itex]

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

does anyone know where my mistake is??
 

Answers and Replies

  • #2
35,135
6,884

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

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

does anyone know where my mistake is??
 
  • #3
35,135
6,884
BTW, this thread was marked as "Solved" but that didn't appear to really be the case, so I have changed it to "Unsolved."
 

Related Threads on Finding convergence of this series using Integral/Comparison

Replies
2
Views
576
Replies
8
Views
908
Replies
1
Views
1K
Replies
6
Views
2K
  • Last Post
Replies
3
Views
553
Replies
15
Views
761
Replies
6
Views
1K
  • Last Post
Replies
14
Views
771
Top