Infinite Series (Integral Test)

[Fernando Rios]

Homework Statement

Use the integral test to find whether the following series converge or diverge.

Relevant Equations

∑n=3∞(1/(n^2-4))

I got the following expression:

-(1/4)ln((n+2)/(n-2))

When I substitute "∞" in the expression I found it undefined. However, the book says the series converges. What am I doing wrong?

 

[BvU]

When I substitute "∞" in the expression

And how do you do such a thing ?

Do you know of a way to rewrite $$\sum_{n=3}^\infty\ {1\over n^2-4}\ ?$$ (hint: write down a few terms ...)

[edit] Oops ! Thanks Math !

 

[member 587159]

Homework Statement: Use the integral test to find whether the following series converge or diverge.
Homework Equations: ∑n=3∞(1/(n^2-4))

I got the following expression:

-(1/4)ln((n+2)/(n-2))

When I substitute "∞" in the expression I found it undefined. However, the book says the series converges. What am I doing wrong?

When you "substitute $\infty$" in $-1/4 \ln((n+2)/(n-2))$ you get $\ln(1) = 0$.

The integral test is probably the easiest way to deduce convergence here (or using results about hyperharmonic series).

 

[Fernando Rios]

Thank you for your answer. Isn't ∞/∞ indeterminate?

 

[BvU]

I found it undefined

How so ? If you rewrite $$ {n+2\over n-2} = {1 + {2\over n} \over 1 - {2\over n} }$$ then you see that this goes to 1 if $n\uparrow\infty$ .

 

[HallsofIvy]

Yes, "$\frac{\infty}{\infty}$" is "indeterminate- but, do you understand what "indeterminate" means? It does not necessarily mean that there is no value. It simply means that the value is not determined directly by that form but might have a value calculated in some other form. That is what BvU did.

 

[Fernando Rios]

Got it. Thank you for your answers.