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More specifically, I'm supposed to compute $\displaystyle\sum_{k=1}^{n} \frac 1 {k(k + 1)} $ by using the equality $\frac 1 {k(k + 1)} = \frac 1 k - \frac 1 {k + 1}$ and the problem before which just says that, $\displaystyle\sum_{j=1}^{n} a_j - a_{j - 1} = a_n - a_0$.

I can add up the sum for any $n$ but I'm not sure what they mean by "compute".


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I take "compute" to mean "write down a formula that works for any $n$ and doesn't have any summation signs or dot-dot-dots in it." –  Gerry Myerson Feb 3 '13 at 1:04
Compute here means express the sum as a function of $n$ that does not involve a summation. –  Brian M. Scott Feb 3 '13 at 1:04
They mean to express answer using only letter "n" and signs "1", "/","-" –  no identity Feb 3 '13 at 1:04

2 Answers 2

up vote 2 down vote accepted

Find a formula in terms of $n$ that can give you the sum for any $n$:

$\displaystyle\sum_{k=1}^{n} \dfrac{1}{k(k + 1)} = \sum_{k=1}^n \dfrac{1}{k} - \dfrac{1}{(k + 1)} = \sum_{k=1}^n\;\left(-\frac{1}{k+1} - \left(\frac{-1}{k}\right) \right) = \;\;\;?$

Knowing that $\displaystyle\sum_{k=1}^{n} a_k - a_{k - 1} = a_n - a_0$, just determine what $a_k$ and $a_{k - 1}$ represent in terms of your sum, and then express the sum in terms of the hint: in terms of $a_n$, and $a_0$ ($a_0$) meaning the $a_k$ term evaluated at k=0).

Edit for clarification:

$$a_k \iff -\dfrac{1}{k+1} \implies a_n = -\dfrac{1}{n+1},\;a_0 = -\dfrac{1}{0 + 1}$$

$$\implies a_n - a_0 = -\dfrac{1}{n+1} - \left(-\dfrac{1}{1}\right) = 1 - \dfrac{1}{n+1} = \dfrac{n}{n+1}$$

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Okay, so that's what I thought it might have meant. I've been struggling though. I don't see how the problem before helps me... –  papercuts Feb 3 '13 at 1:18
+1 I think the summation exactly before "?" needs a parentheses. (-1/k+1--1/k), I mean. –  Babak S. Feb 3 '13 at 3:25
thanks for the tip...I added them to ensure it's understood the entire "difference" is the summand! –  amWhy Feb 4 '13 at 3:41

It means: find how much it sums to. In fact, you have already said everything you need to solve the problem. You only have to put 1 and 1 together to obtain 2.

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