# Integral test for converging series

1. May 12, 2009

### lha08

1. The problem statement, all variables and given/known data
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?

2. Relevant equations

3. The attempt at a solution

2. May 12, 2009

### gabbagabbahey

Re: Series

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)

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. May 12, 2009

### lha08

Re: Series

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...

4. May 12, 2009

### gabbagabbahey

Re: Series

There is no way this series converges for n=1 to n=infinty.... the problem is the n=5 term...

$$f(5)=1/(5^2-4(5)-5)=1/0=\infty$$

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

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