# Determine the convergence or divergence of the infinite series

1. Apr 13, 2014

### QuantumCurt

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

This is for Calculus II. We've just started the chapter on Infinite Series. n runs from 1 to ∞.

$$\Sigma\frac{1}{n(n+3)}$$

3. The attempt at a solution

I used partial fraction decomposition to rewrite the sum.

$$\frac{1}{n(n+3)}=\frac{A}{n}+\frac{B}{n+3}$$

$$1=A(n+3)+Bn$$

Letting n=0, and then letting n=-3, I get -

$$\frac{1}{n(n+3)}=\frac{1/3}{n}-\frac{1/3}{n+3}$$

Then I factored the 1/3 out, and rewrote the sum.

$$\frac{1}{3}\Sigma \ [\frac{1}{n}-\frac{1}{n+3}]$$

I know this is going to be a telescoping series, but my problem is that I can't seem to figure out how to write an expression for the nth term. Writing the first few terms of the series gets me to -

$$\frac{1}{3}[(1-1/4)+(1/2-1/5)+(1/3-1/6)+(1/4-1/7)+....]$$

None of the terms seem to be canceling. I wrote the series out several terms farther than this, and I still wasn't getting any cancellation.

My solutions manual is showing the expression being written as follows -

$$\frac{1}{3}[(1-1/4)+(1/2-1/5)+(1/3-1/6)+(1/4-1/7)+....+(\frac{1}{n-2}-\frac{1}{n+1})-(\frac{1}{n-1}-\frac{1}{n+2})-(\frac{1}{n}-\frac{1}{n+3})]$$

Which they then rewrite as -

$$\frac{1}{3}[1+\frac{1}{2}+\frac{1}{3}-\frac{1}{n+1}-\frac{1}{n+2}-\frac{1}{n+3}]$$

The final solution after taking the limit is $\frac{11}{8}$

I'm not seeing the intuition behind getting to this expression. Anyone input would be much appreciated.

2. Apr 13, 2014

### Dick

I already see the +1/4 that will cancel the -1/4 in your series. If you write out the next term you'll find a +1/5. There certainly is cancellation between numbers that are three terms apart.

Last edited: Apr 13, 2014
3. Apr 13, 2014

### QuantumCurt

Wow, I must be more tired than I thought. I have no idea how that went right past me. Thanks.

I'm getting cancellation now for the second part of the first three terms, and for the first part of every following term.

I'm still not seeing how they're getting the nth term expression though.

4. Apr 13, 2014

### Dick

Write out some more last terms in your series. It's the same game. The first three positive numbers in the terms of your series won't cancel and neither will the last three negative ones. Like I said, the cancellations are going to be three terms apart. You see why, right?

5. Apr 13, 2014

### QuantumCurt

Okay, I think I'm seeing it now. I wasn't writing the series out far enough.

$$\frac{1}{3}[(1-1/4)+(1/2-1/5)+(1/3-1/6)+(1/4-1/7)+(1/5-1/8)+(1/6-1/9)+(1/7-1/9)+...$$

Now I can see how all of the terms are going to cancel aside from the positive part of the first three, and the negative part of the last three.

For the last part...

$$\frac{1}{3}[(1-1/4)+(1/2-1/5)+(1/3-1/6)+(1/4-1/7)+(1/5-1/8)+(1/6-1/9)+(1/7-1/9)+...$$

$$...+(\frac{1}{n-2}-\frac{1}{n+1})-(\frac{1}{n-1}-\frac{1}{n+2})-(\frac{1}{n}-\frac{1}{n+3})]$$

The first part is 1/(n-2)-1/(n+1) simply because it's two terms before the last term...right? Same goes for the second part. Why are the last two terms negative though? That's the last thing I'm not getting here.

6. Apr 13, 2014

### Dick

Same problem at the end. You aren't writing out enough terms. The one before the last one you wrote is 1/(n-3)-1/n. That -1/n will cancel the 1/n at then end. Hey, why did the + signs between terms change into - signs at the end?

7. Apr 13, 2014

### QuantumCurt

Okay...so -

$$\frac{1}{3}[(1-1/4)+(1/2-1/5)+(1/3-1/6)+(1/4-1/7)+(1/5-1/8)+(1/6-1/9)+(1/7-1/9)+...$$

$$...+(\frac{1}{n-3}-\frac{1}{n})+(\frac{1}{n-2}-\frac{1}{n+1})-(\frac{1}{n-1}-\frac{1}{n+2})-(\frac{1}{n}-\frac{1}{n+3})]$$

I'm still not getting why those terms at the end are negative though. Is that a typo in the solutions manual? It makes more sense for them to be positive...like so

$$\frac{1}{3}[(1-1/4)+(1/2-1/5)+(1/3-1/6)+(1/4-1/7)+(1/5-1/8)+(1/6-1/9)+(1/7-1/9)+...$$

$$...+(\frac{1}{n-3}-\frac{1}{n})+(\frac{1}{n-2}-\frac{1}{n+1})+(\frac{1}{n-1}-\frac{1}{n+2})+(\frac{1}{n}-\frac{1}{n+3})]$$

If they're written as being negative, the 1/n's won't cancel. There's another typo in the manual...it says the final solution is 11/8, but it should be 11/18

8. Apr 13, 2014

### QuantumCurt

Then all of this leads me to -

$$S=\frac{1}{3}[1+\frac{1}{2}+\frac{1}{3}-\frac{1}{n+1}-\frac{1}{n+2}-\frac{1}{n+3}]$$

Then I take the limit as n→∞ and I get $\frac{11}{18}$

9. Apr 13, 2014

### lurflurf

you can manipulate the index

$$\sum_{n=1}^\infty \frac{1}{n(n+3)}=\frac{1}{3}\left(\sum_{n=1}^\infty \frac{1}{n}-\sum_{n=1}^\infty \frac{1}{(n+3)}\right)=\lim_{N\rightarrow\infty} \frac{1}{3}\left(\sum_{n=1}^3 \frac{1}{n}+\sum_{n=4}^N \frac{1}{n}-\sum_{n=1}^{N-3} \frac{1}{(n+3)}-\sum_{n=N-2}^{N} \frac{1}{(n+3)}\right)$$

or think of three sacks
person A puts {1,2,3} in the plus sack
and {4,5,6,...,N} in the zero sack
person B puts {4,5,6,...,N} in the zero sack
then puts in {N+1,N+2,N+3} in the minus sack

What are in the plus and minus sacks?
We do not care about the zero sack.

10. Apr 13, 2014

### lurflurf

yes that is right

11. Apr 13, 2014

### Dick

Yes, I'm sure it's a typo.

12. Apr 13, 2014

### QuantumCurt

We just started this chapter, so I haven't seen anything like this yet. I imagine we'll be getting to this kind of stuff soon enough though.

Okay, I thought it had to be. That was really throwing me off.

Thanks for all the help...I've got it now.

13. Apr 13, 2014

### Ray Vickson

What was your question: (1) Find the sum of the series; or (2) Determine if the series converges? Of course, you can do (2) by answering (1), but if all you really want to do is answer (2), it would be much easier to use a simple convergence test:
$$0 < t_n \equiv \frac{1}{n(n+3)} < \frac{1}{n^2},$$
and we know already that $\sum 1/n^2$ converges.

14. Apr 13, 2014

### QuantumCurt

The question simply stated "Determine the convergence of divergence of the series."