Convergence: The Integral Test

[Joshk80k]

Homework Statement

Is

\sum \frac{1}{2n(2n+1)}

convergent or divergent?

(Note that the summation is from 1 to infinity)

Homework Equations

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

IF
L = \infty, divergent
L < \infty, 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,

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

\int \frac{1}{2n(2n+1)}dx = \int \frac{A}{2n} + \frac{B}{2n+1}

1 = 2nA + 2nB + A

Matching up coefficients,

A^0: 1 = A

and

A^1: 0 = 2A + 2B,

B = -A = -1

\int \frac{1}{2n(2n+1)}dx = \int \frac{1}{2n} + \frac{-1}{2n+1}

Using substitution,

\int \frac{1}{2n} = \frac{1}{2}ln(2n)

and

\int \frac{-1}{2n+1} = -\frac{1}{2}ln(2n+1)

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

\frac{1}{2}ln(2n) + -\frac{1}{2}ln(2n+1) = \frac{1}{2}ln(\frac{2n}{2n+1})

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

\frac{1}{2}ln(\frac{2t}{2t+1}) - \frac{1}{2}ln(\frac{2}{3})

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 \sum 1/(4n^2).

 

[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?

\frac{1}{\sqrt{n(n+1)}}

to \frac{1}{\sqrt{n(n)}} = \frac{1}{n}.

 

[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?

\frac{1}{\sqrt{n(n+1)}}

to \frac{1}{\sqrt{n(n)}} = \frac{1}{n}.

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
\frac{1}{\sqrt{n(n+2)}}

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.

\frac{1}{2n(2n+1)} = \frac{1}{2n}-\frac{1}{2n+1}

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