What does it mean to "compute" a generic formula without values?

Question

Very elementary math question here in regards to my discrete math class. I've got a problem here that says...

"Compute the following":

$$\sum_{j=1}^n \frac{1}{j(j+1)}$$

What on earth does it mean to "compute" this? I mean, without any values, does that just mean to find another way of re-arranging the components to get a formula of equal value, just in a different form?

I did some iterations by hand and found that for n=4, we end up with:

$$\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}$$

So it looks like each iteration just increases the denominator by the previous increase's next even number:

$\frac{1}{2} \rightarrow \frac{1}{6}$ (increase by 4)

$\frac{1}{6} \rightarrow \frac{1}{12}$ (increase by 6)

$\frac{1}{12} \rightarrow \frac{1}{20}$ (increase by 8)

So I see the pattern. Does that mean for example that I could re-write the above sigma summation as a "rule" for making the sequence?

Answer 1

You’re supposed to find a closed form expressing that sum as a function of $n$, one that does not include a summation. An example that you’ve probably seen is the formula for the sum of the first $n$ positive integers:

$$\sum_{k=1}^nk=\frac{n(n+1)}2\;.$$

You’re supposed to do something similar here. Here’s a hint to get you started:

$$\frac1{k(k+1)}=\frac1k-\frac1{k+1}\;.$$

Answer 2

You are not given a specific value like $1,2,68$ but your summation depends $n$ , hence your result will be a function of $n$. It's always a good idea to get your hands dirty and look for patterns but in this case there is a genral rule:

If you have a summation with a polynomial wich can be factorized at the denomnator ( in this case: $j(j+1)=j^2+j \ \ $) it is a good idea to split the terms in $l$ parts where $l$ is the number of factors of the polynomial. In our case $l=2$ as you can see.

$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \dfrac1{k(k+1)}=\dfrac{A}{k}+\dfrac{B}{k+1} \ $

With a little computation you will find out : $A=1$ and $B=-1$

now do you see a pattern here ?