Integral test for converging series

  • Thread starter lha08
  • Start date
  • #1
164
0

Homework Statement


In one of my problems, i'm asked to use integral test in order to determine whether the following function is converging or diverging:
(summation from n=1 to inf) 1/(n^2-4n-5)
I was just wondering how do we know if function is positive, continuous and decreasing because when i tried to see if it was positive (i plugged in a value), i got a negative...
Also, for a integral test, does the entire function have to be decreasing?


Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
gabbagabbahey
Homework Helper
Gold Member
5,002
6


Homework Statement


In one of my problems, i'm asked to use integral test in order to determine whether the following function is converging or diverging:
(summation from n=1 to inf) 1/(n^2-4n-5)
I was just wondering how do we know if function is positive, continuous and decreasing because when i tried to see if it was positive (i plugged in a value), i got a negative...
Also, for a integral test, does the entire function have to be decreasing?
Is the summation really form n=1 to infinity? If so, I can tell you without testing that the series diverges (take a look at the the n=5 term:wink:)

Anyways, for the integral test to be a valid test, your summand (function) must be monotone decreasing and non-negative on the entire summation interval. If you plug in a value of 'n' (on the summation interval) and get a negative number you should know right there that the integral test is useless.
 
  • #3
164
0


Is the summation really form n=1 to infinity? If so, I can tell you without testing that the series diverges (take a look at the the n=5 term:wink:)

Anyways, for the integral test to be a valid test, your summand (function) must be monotone decreasing and non-negative on the entire summation interval. If you plug in a value of 'n' (on the summation interval) and get a negative number you should know right there that the integral test is useless.
You said that's it's apparently diverging, but in the example, we found it to be converging:(-1/6)ln(-2)...i dunno 'cause you mentioned that it has to be non-negative on its entire summation interval...:confused:
 
  • #4
gabbagabbahey
Homework Helper
Gold Member
5,002
6


You said that's it's apparently diverging, but in the example, we found it to be converging:(-1/6)ln(-2)...i dunno 'cause you mentioned that it has to be non-negative on its entire summation interval...:confused:
There is no way this series converges for n=1 to n=infinty.... the problem is the n=5 term...

[tex]f(5)=1/(5^2-4(5)-5)=1/0=\infty[/tex]

How can a sum converge when one of its terms is undefined?!

Also, [tex]\int_1^\infty 1/(x^2-4x-5)dx\neq\frac{-1}{6}\ln(-2)[/tex] ...the integral doesn't even converge!
 

Related Threads on Integral test for converging series

  • Last Post
Replies
1
Views
1K
  • Last Post
2
Replies
28
Views
3K
Replies
1
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
847
  • Last Post
Replies
2
Views
987
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
6
Views
1K
Replies
4
Views
665
Replies
1
Views
803
Top