Most likely simple Math question 'bout sums

[/\/\isanThr0p]

Hey everyone!

Well here's my question:
I need to prove that
Sum(i=1,n,(1/(i(i+1)))=n/(n+1)

I know that it should be pretty simple but right now I don't know where to start, since I am not sure what to do with variables in the denominator.

thanks in advance
....(I haven't postet here for ages... hope everything's still the same)

[/\/\isanThr0p]

I hope the notation is alright for you, I just asked a friend over ICQ and he didn't get it, so I came back to explain. I hope noone considers this spamming, since this question was on top anyways.

well what I meant is the Sum (capital greek epsilon) over the formula (whicht should be clear) from i=1 to i=n

[TheManWhoCan]

First you need to split it into partial fractions (I'm gonna use "r" so I don't get confused), so:

1/(r(r+1)) = 1/r - 1/(r+1)

Now write Sum(thing) =

1/1 - 1/2 (that's putting r=1)
+1/2 - 1/3 (r=2)
+1/3 - 1/4
etc...
+1/(n-1) - 1/n (r=n-1)
+1/n - 1/(n+1) (r=n)

Notice that for each line, the first term will cancel with the second on the line before, so the -1/2 cancels with the 1/2 etc... However for the first line, there is no line before, so the first term on the first line doesn't cancel. Also after the last line there are no more lines (!) so the second term of the last line can't cancel.

So the answer is whatever's left:

1 - 1/(n+1)

[/\/\isanThr0p]

Thanks a lot!
By seperating the term, it was easy to prove.

There is also a way to do it with sums so I don't have to put "...". But I am pretty sure you could come up with that as easily as you did with the solution you posted before.

Thanks again

[TheManWhoCan]

Well the other common way of doing finite sums is to put the thing in terms of powers of r: such as r, r², r³, etc, the formulae for the sum from 1 to n of which are known, or runs of r: r, r(r+1), r(r+1)(r+1), etc, for which the formulae are also known. I can't see how you'd apply that here though.