Sum of the series from n=3 to infinity of 1/((2n-3)(2n-1))

[c.dube]

Homework Statement

Find the sum of the series:
$$\sum^{\infty}_{n=3}\frac{1}{(2n-3)(2n-1)}$$

Homework Equations

N/A

The Attempt at a Solution

This isn't geometric, I can't get it from any common Maclaurin series (as far as I can work out). The book I have tells me the answer is 1/6, I'm sure I'm doing something stupid. Thanks in advance for any help!

 

[Mark44]

Try breaking up the summand using partial fraction decomposition, and then expand the series.

 

[c.dube]

OK, so I have $$\sum^{\infty}_{n=3}\frac{1}{2(2n-3)}-\frac{1}{2(2n-1)}$$, but I'm not sure what you mean by "expanding the series"; should I split it into two summations? And if so, where do I go from there?

 

[Mark44]

No, not two series - just one. Expanding the series means writing out the sum of terms, starting with the one for n = 3. Look at the sequence of partial sums, S_k, the sum of the terms from n = 3 to n = k.

 

[c.dube]

*hits self in head* Wow! Thanks a ton, that should have been obvious.

 

[Mark44]

These things are obvious only after you have done them a time or two.

 

[c.dube]

Yah, well at least now when I see it again I'll know what to do! Thanks again for the help.