Convergent/Divergent Problem

[thatguy47]

Determine whether the series is convergent or divergent. If it is convergent, find its sum.
Here's the problem:


Here's the solution to the problem (the image is to big to post):
http://img398.imageshack.us/img398/4150/math2xb9.jpg

I understand it up until the: 1/2 + 1/3 - 1/(n+2) - 1/(n+3) >>>>How do you get that? Can you explain how the problem is solved from that step forward. Thanks in advance.

 

[galactus]

Look close. Those are the only terms that don't cancel out.

 

[fasteddie65]

This type of series is sometimes called an accordion series, since everything cancels except the first and last terms.

 

[thatguy47]

I see now how the terms cancel out but what about the last ones:

How do you get

from that? What I'm trying to say is how does the (1/n) and (1/(n+1)) cancel out?

 

[PAULK]

I see now how the terms cancel out but what about the last ones:

How do you get

from that? What I'm trying to say is how does the (1/n) and (1/(n+1)) cancel out?

The way to understand it is to understand EXACTLY what the problem says:

SUM[j = 1 to infinity] (something) MEANS:

lim(n --> infinity) SUM[j = 1 to infinity] (something)

So write the expression as a sum going to the n-th term. That will be your sum having four terms, because all terms that have denominators smaller than n+2 and n+3 have cancelled, except for the first two. It looks to me that

If you are having difficulty grasping the concept, try writing it out for n = 5, say. Or n = 8. See exactly which terms survive and why.