Series convergence

1. Jun 10, 2009

utopiaNow

1. The problem statement, all variables and given/known data
Use partial fractions to show
$$\displaystyle\sum_{n=1}^\infty \frac{1}{n(n+1)(n+2)} = \frac{1}{4}$$

3. The attempt at a solution
I did the partial fraction decomposition to get: $$\displaystyle\sum_{n=1}^\infty \frac{1}{2n} - \frac{1}{n + 1} + \frac{1}{2n + 4}$$

I'm not sure how to proceed from here, in my textbook the example shows how terms in the partials sums overlap and cancel out if you start looking at the terms in the partial fraction decomposition, however I can't see that happening with this particular series.

Any suggestions would be appreciated. Thanks in advance.

2. Jun 10, 2009

rock.freak667

Write out a few term, n=1,2,3,4,5....N-3,N-2,N-1,N

then check as N->inf.

3. Jun 10, 2009

utopiaNow

Hi,

I tried doing that originally, but I don't see any patterns that I can exploit when I write the terms of the partial decomposition for n = 1, 2, 3,..., N-2, N-1, N as you stated.

Do you have any further suggestions?

4. Jun 10, 2009

Dick

Write the terms for successive values of n each on a row 1/(2n+4) -1/(n+1) 1/(2n)
1/6 -1/2 1/2
1/8 -1/3 1/4
1/10 -1/4 1/6
1/12 -1/5 1/8
1/14 -1/6 1/10

Oblique hint: diagonal.

5. Jun 10, 2009

utopiaNow

Interesting, so everything seems to be canceling out if you keep going long enough diagonally, except the 1/4 in the second row. However is there any way to formalize this? Or is simply noticing this pattern enough for a formal proof?

6. Jun 10, 2009

Dick

Depends on how formal you want to be. Writing a table like that and scratching out the cancellations is good for me. If you want to do it formally write out the first term from the n case, the second term from the n+1 case and the third term from the n+2 case and show they cancel algebraically. What is important is to realize just because there isn't an obvious cancellation doesn't mean there isn't one.

7. Jun 10, 2009

utopiaNow

Thank you for the help!

8. Jun 10, 2009

txy

why not take out 1/2 from the partial fraction decomposition so that it's more obvious?

$$\sum^{+\infty}_{n=1}(\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{2n+4})$$
$$= \frac{1}{2}\sum^{+\infty}_{n=1}(\frac{1}{n}-\frac{2}{n+1}+\frac{1}{n+2})$$
$$= \frac{1}{2}\sum^{+\infty}_{n=1}(\frac{1}{n}-\frac{1}{n+1}+\frac{1}{n+2}-\frac{1}{n+1})$$
$$= \frac{1}{2}\sum^{+\infty}_{n=1}(\frac{1}{n}-\frac{1}{n+1}) + \frac{1}{2}\sum^{+\infty}_{n=1}(\frac{1}{n+2}-\frac{1}{n+1})$$

9. Jun 10, 2009

Dick

That's a nice way to handle it. There's more than one way to skin a cat.