Convergence Series: Partial Fractions Homework

[utopiaNow]

Homework Statement

Use partial fractions to show
\displaystyle\sum_{n=1}^\infty \frac{1}{n(n+1)(n+2)} = \frac{1}{4}

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.

 

[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.

 

[utopiaNow]

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

then check as N->inf.

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?

 

[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.

 

[utopiaNow]

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.

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?

 

[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.

 

[utopiaNow]

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.

Thank you for the help!

 

[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})<br />

 

[Dick]

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