Convergence: The Integral Test

[Joshk80k]

Homework Statement

Is

[tex]\sum \frac{1}{2n(2n+1)}[/tex]

convergent or divergent?

(Note that the summation is from 1 to infinity)

Homework Equations

[tex]\int f(x) dx = L[/tex], (range is from 1 to infinity)

IF
L = [tex]\infty[/tex], divergent
L < [tex]\infty[/tex], convergent.

The Attempt at a Solution

I tried a number of tests, and this is the only convergence test that I am not sure about. I attempted the ratio and root tests, but both were inconclusive.

As for my attempt,

[tex]\int \frac{1}{2n(2n+1)}dx[/tex], using the partial fractions method to integrate,

[tex]\int \frac{1}{2n(2n+1)}dx = \int \frac{A}{2n} + \frac{B}{2n+1} [/tex]

[tex]1 = 2nA + 2nB + A[/tex]

Matching up coefficients,

[tex]A^0: 1 = A[/tex]

and

[tex]A^1: 0 = 2A + 2B,[/tex]

[tex]B = -A = -1[/tex]

[tex]\int \frac{1}{2n(2n+1)}dx = \int \frac{1}{2n} + \frac{-1}{2n+1} [/tex]

Using substitution,

[tex]\int \frac{1}{2n} = \frac{1}{2}ln(2n)[/tex]

and

[tex]\int \frac{-1}{2n+1} = -\frac{1}{2}ln(2n+1)[/tex]

So, adding these two together, and using properties of the natural log,

[tex]\frac{1}{2}ln(2n) + -\frac{1}{2}ln(2n+1) = \frac{1}{2}ln(\frac{2n}{2n+1})[/tex]

Now, inserting the bounds 1 and t (Where t is infinity),

[tex]\frac{1}{2}ln(\frac{2t}{2t+1}) - \frac{1}{2}ln(\frac{2}{3})[/tex]

Here is where I am a little stumped. I want to say "Hey, t is infinity, and at large values of t, we can ignore the excess numbers (In this case, 2t + 1 is just a tiny bit different than 2t, so we ignore the 1 to make the cancellation) and as a result, I'll have a finite value, and for this reason, the answer is convergent.

BUT, I have always been a little shaky on convergence tests - am I doing this correctly or am I wrong to make that assumption about the cancellation?

 

[Mark44]

An easier test for this one is comparison with the convergent series [itex]\sum 1/(4n^2)[/itex].

 

[Joshk80k]

Sigh.

Using your method, I got the answer in 15 seconds, as opposed to the 15 minutes my other way took.

Thanks to you, I know this is now convergent - but, for future reference, is it OK for me to make that assumption that I made towards the end of the problem?

 

[Mark44]

Yes. Adding or subtracting a constant won't affect the decision of convergence/divergence. If the integral diverges (to infinity), adding or subtracting a little bit doesn't change that. Same thing if the integral converges. Subtracting or adding something just gives a different (finite) number.

 

[Joshk80k]

Ah OK, thanks so much for your help! My professor gave us a whole list of these things to figure out, and if I'm applying the comparison test you just showed me correctly, then I shouldn't have any more problems.

Am I applying what you just showed me in the correct manner here?

[tex] \frac{1}{\sqrt{n(n+1)}} [/tex]

to [tex] \frac{1}{\sqrt{n(n)}} = \frac{1}{n}[/tex].

 

[Dick]

Ah OK, thanks so much for your help! My professor gave us a whole list of these things to figure out, and if I'm applying the comparison test you just showed me correctly, then I shouldn't have any more problems.

Am I applying what you just showed me in the correct manner here?

[tex] \frac{1}{\sqrt{n(n+1)}} [/tex]

to [tex] \frac{1}{\sqrt{n(n)}} = \frac{1}{n}[/tex].

You've got 1/(sqrt(n*(n+1))<=1/sqrt(n*n)=1/n. You know 1/n is divergent, right? If so that's not good enough. You shown your original series is less than a divergent series. That doesn't prove anything. If you want to prove it diverges you need to show it's GREATER than a divergent series. Any ideas?

 

[Joshk80k]

I want to relate it to something like
[tex]\frac{1}{\sqrt{n(n+2)}}[/tex]

But to be honest I can't think of an easy way to deal with that one either...I'll continue thinking about it...thanks for correcting it though =).

 

[Dick]

How about comparing 1/sqrt(n*(n+1)) with 1/sqrt(n*(n+n))? Which is greater and what can you say about the convergence of the second without working too hard?

 

[Joshk80k]

I can say that my original function is greater than this new function, and that this new function is also divergent.

OK, I can see how your choice coincides with the way the convergence test is written. I just made the mistake of jumping the gun with this new technique =).

 

[Dick]

I can say that my original function is greater than this new function, and that this new function is also divergent.

OK, I can see how your choice coincides with the way the convergence test is written. I just made the mistake of jumping the gun with this new technique =).

I think you are getting it. You can always make convenient choices in the comparison to make them easier to evaluate.

 

[rs1n]

Sigh.

Using your method, I got the answer in 15 seconds, as opposed to the 15 minutes my other way took.

Thanks to you, I know this is now convergent - but, for future reference, is it OK for me to make that assumption that I made towards the end of the problem?

Your attempt is not a complete loss. Actually, there is something interesting to gain: the notion of telescoping series.

[tex]\frac{1}{2n(2n+1)} = \frac{1}{2n}-\frac{1}{2n+1}[/tex]

Within your sum, replace [itex]\frac{1}{2n(2n+1)}[/itex] with [itex]\frac{1}{2n}-\frac{1}{2n+1}[/itex]. Then write out a few terms, and observe any simplifications via grouping...