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:


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?

  • 1
    $\begingroup$ I think your teacher means to say:express the sum solely in terms of $n$. $\endgroup$
    – RKD
    Jul 23, 2013 at 2:19
  • 2
    $\begingroup$ Write the result as an expression depending on $n$. $n = 1 \leadsto \frac12$, $n = 2 \leadsto \frac23$, $n = 3 \leadsto \frac34$. See a pattern? $\endgroup$ Jul 23, 2013 at 2:20
  • $\begingroup$ Yep, you can rearrange it. Try writing the sum in the form $\frac{A}{j}+\frac{B}{j+1}$. Figure out A and B, then you'll see that the sum telescopes $\endgroup$ Jul 23, 2013 at 2:21

2 Answers 2


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:


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



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 ?


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