Work for how Σ∞(top)n=1(bottom) 1/(n^2-4n-5) converges after n=5 (Integral Test)?

Question

I have asked before but still am having problems showing all the necessary work to prove this.

Okay I got to 1/6 lim ln(x-5/x+1) | (1 bottom, infinity top). What do I do now.

Answer 1

for computing the integral, use that
1/(n^2-4n-5) = 1/(n-5)(n+1) = 1/6(1/(n-5) - 1/(n+1)).

Okay I got to 1/6 lim ln(x-5/x+1) | (1 bottom, infinity top). What do I do now.

Actually it's 6 at bottom !

1/6 (lim( ln (x-5)/(x+1)) -1/6 ln(6-5)/(6+1) =
= 0 - 1/6 ln(1/7) = 1/6 ln 7. Since the integral is finite, the series converges.

Answer 2

[I didn't know how far you got] The integral test starts with converting f(n) into f(x). In this case, f(x)=1/(x^2-4x-5) Then you take the integral of this from x=1 to x=infinity. The function f(x) should be factored on the bottom so that f(x)=1/[(x-5)(x+1)] and right away you should see an integral that is calculated by splitting the fraction into 1/[(x-5)(x+1)]=A/(x-5) + B/(x+1). To figure what A and B are, all you have to do is set the equation A(x+1) + B(x-5)=1. This can be accomplished by saying that A+B=0 and A-5B=1. A=1/6 and B=-1/6. Therefore f(x)=1/[6(x-5)] - 1/[6(x+1)].

This integral should be relatively straight forward. Using substitutions. The trick is that when you are using infinity in the definite integral, that you don't just substitute infinity into the equation, but you have to turn the equation into a limit of sum variable as that variable approaches infinity. The ultimate term that will determine whether the integral converges is not is the lim. as z approaches infinity of 1/6ln(z)-1/6ln(z), which is a L'Hospital indeterminate difference. This is solved simply by factoring ln(z) so that f(z)=ln(z) and g(z)=1/6-1/6, where as z approaches infinity, f(z) approaches infinity and g(z) approaches 0. This can be turned into a useful form by transforming f*g into f/(1/g) or ln(z)/(1/0). Now we can differentiate top and bottom. The bottom is just zero. This may not be obvious but the number 1/0 is constant, although undefined, and the derivative of a constant is zero. The top become 1/z, so the limit of the whole function is zero, the limit converges, the integral converges, so the sumation converges.

Answer 3

To clasify the serie S(1,+infinity) 1/(n^2-4n-5) you don´t need the integral test.
Simply compare it to the serie S(1,inf) 1/n^2
The limit of 1/n^2/(1/n^2-4n-5) = lim (1-4/n-5/n^2) = 1 So your serie is of the class as 1/n^2 known as convergent