Question

Find the sum of the series:

Answer 1

\[\sum_{n=1}^{\infty}\frac{ 6 }{ n(n+3) }\]

Answer 2

I only know how to show that it converges. Finding it's actual sum? No idea.

Answer 3

I think you might be able to deconstruct this by using partial fractions into two separate sums that you can then individually find the sum of easier. That would be my strategy, I'm sure there are other methods though.

Answer 4

I've done that

\[\frac{ 2 }{ n }-\frac{ 2 }{ n+3 }\]
Not exactly sure what to do with that, though.

Answer 5

It looks like you might have a telescoping series, do you know how to deal with them?

Answer 6

Well, don't they have to have the same terms back to back or something like that?
\[(1-\frac{ 1 }{ 2 }) + (\frac{1}{2}-\frac{1}{3}) . . . \]

Answer 7

They don't have to be directly back to back.. there could be a "gap" between terms but eventually gets cancelled out

Answer 8

Yeah, although if you think about it, these won't be exactly back to back, more like separated by about 3 terms, since 2/n is 3 steps away from 2/(n+3) right? So start to write out the first 4 or 5 terms and you'll see the pattern I think. =D

Answer 9

Right, but even then I'm not sure, lol. I came up with something along the lines of:

\[3(\frac{1}{2} + \frac{1}{5} + \frac{1}{9} + \frac{1}{14})\]So yeah, pattern, but I guess my brain still isn't catching something here. Even if I could express that pattern as a series just....er, yeah.

Answer 10

Or was I not supposed to do that and keep it in terms of n?

Answer 11

No you were supposed to do that, I would just write out the sum as something like this:

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

Answer 12

Now if you have that written out you can cross out the -2/4 with the +2/4 a little later on. Then you'll see the 2/5 terms drop out and eventually they'll all cross out except a few first terms.

Answer 13

Ah, gotcha. So is this pretty much always the strategy of something that looks telescoping?

Answer 14

More or less, it's actually been over a year since I've taken cal 2 so I just sort of got lucky, but often times it just boils down to using common sense coupled with a little experience. For instance, I left the 2 in there, but from the get-go I'd suggest simplifying it as much as possible by taking the 6 out of the summation since 6/3+6/4+6/5+... is the same as 6*(1/3+1/4+1/5+...) just as an example. If you can turn it into two separate sums that will make it easier to notice too.

In this case you have essentially a sum like 1/n and you're subtracting 1/(n+3), but it isn't too bad to see that 1/n will become 1/(n+3) very shortly, so you could just add up 1/n+1/(n+1)+1/(n+2) and be done with it since after that 1/n+3 will be subtracted away.

Answer 15

Hmm...alrighty, cool, thanks.