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- Thread starter Erfan
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This looks like a telescopic series to me...Prove that 1/(1*2*3) + 1/(2*3*4) + ... + 1/(n*(n+1)*(n+2)) = 1/4 - 1/(2*(n+1)*(n+2)

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Partial fractions !Hello Erfan,

Are you expected to use partial fraction decomposition or induction, or is the choice of method up to you? Can you show us what you have tried so far?

I tried this 1/(r*(r+1)*(r+2)) = 1/(2r) - 1/(r+1) + 1/(2(r+2))

but it seems that this doesn't work

- Feb 13, 2012

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It is convenient to proceed by induction. Let's suppose that...Prove that 1/(1*2*3) + 1/(2*3*4) + ... + 1/(n*(n+1)*(n+2)) = 1/4 - 1/(2*(n+1)*(n+2)

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$\displaystyle S_{n}= \frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 4} + ... + \frac{1}{n \cdot (n+1) \cdot (n+2)} = \frac{1}{4} - \frac{1}{2 \cdot (n+1) \cdot (n+2)}\ (1)$

... is true for some n [it is true for n=2...]. Then is...

$\displaystyle S_{n+1} = \frac{1}{4} - \frac{1}{2 \cdot (n+1) \cdot (n+2)} + \frac{1}{(n+1) \cdot (n+2) \cdot (n+3)} = \frac{1}{4} - \frac{1}{2 \cdot (n+2) \cdot (n+3)}\ (2)$

... and that proves the statement...

Kind regards

$\chi$ $\sigma$

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- Feb 7, 2012

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Yes it does work! Try it like this: $$\begin{array}{cccccc} S_n = \frac{1/2}1 &- \frac12 &+ \frac{1/2}3 \\ & + \frac{1/2}2 &- \frac13 &+ \frac{1/2}4 \\ && + \frac{1/2}3 &- \frac14 &+ \frac{1/2}5 \\ &&& + \frac{1/2}4 &-\frac15 &+\frac{1/2}6 \\ &&&& + \ldots, \end{array}$$ continuing like that until you get to the row $+\dfrac{1/2}n - \dfrac1{n+1} + \dfrac{1/2}{n+2}.$ Notice that apart from a few terms at the beginning and end of the sum, the terms in each column add up to $0$.Partial fractions !

I tried this 1/(r*(r+1)*(r+2)) = 1/(2r) - 1/(r+1) + 1/(2(r+2))

but it seems that this doesn't work

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\(\displaystyle \sum_{k=1}^n\left(\frac{1}{k(k+1)(k+2)} \right)\)

Now, using the Heaviside cover-up method on the summand:

\(\displaystyle \frac{1}{k(k+1)(k+2)}=\frac{A}{k}+\frac{B}{k+1}+ \frac{C}{k+2}\)

\(\displaystyle A=\frac{1}{2},\,B=-1,\,C=\frac{1}{2}\)

You had the correct decomposition. Thus, our sum may be written:

\(\displaystyle \sum_{k=1}^n\left(\frac{1}{2k}-\frac{1}{k+1}+\frac{1}{2(k+2)} \right)\)

\(\displaystyle \frac{1}{2}\sum_{k=1}^n\left(\frac{1}{k} \right)-\sum_{k=1}^n\left(\frac{1}{k+1} \right)+\frac{1}{2}\sum_{k=1}^n\left(\frac{1}{k+2} \right)\)

I would split the middle sum into two halves and group as follows:

\(\displaystyle \frac{1}{2}\left(\left(\sum_{k=1}^n\left(\frac{1}{k} \right)-\sum_{k=1}^n\left(\frac{1}{k+1} \right) \right)+ \left(\sum_{k=1}^n\left(\frac{1}{k+2} \right)-\sum_{k=1}^n\left(\frac{1}{k+1} \right) \right) \right)\)

Now, re-indexing the first sum in each group, we may write:

\(\displaystyle \frac{1}{2}\left(\left(\sum_{k=0}^{n-1}\left(\frac{1}{k+1} \right)-\sum_{k=1}^n\left(\frac{1}{k+1} \right) \right)+ \left(\sum_{k=2}^{n+1}\left(\frac{1}{k+1} \right)-\sum_{k=1}^n\left(\frac{1}{k+1} \right) \right) \right)\)

We want to get the sums to cancel, so let's look at the first group:

\(\displaystyle \sum_{k=0}^{n-1}\left(\frac{1}{k+1} \right)-\sum_{k=1}^n\left(\frac{1}{k+1} \right)\)

Pulling of the first term from the first sum and the last term from the second, we have:

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

And so we are left with:

\(\displaystyle 1-\frac{1}{n+1}\)

Let's next look at the second group:

\(\displaystyle \sum_{k=2}^{n+1}\left(\frac{1}{k+1} \right)-\sum_{k=1}^n\left(\frac{1}{k+1} \right)\)

Pulling off the last term from the first sum and the first term from the second, we have:

\(\displaystyle \sum_{k=2}^{n}\left(\frac{1}{k+1} \right)+\frac{1}{n+2}-\frac{1}{2}-\sum_{k=2}^n\left(\frac{1}{k+1} \right)\)

And so we are left with:

\(\displaystyle \frac{1}{n+2}-\frac{1}{2}\)

Our sum may then be written as:

\(\displaystyle \frac{1}{2}\left(1-\frac{1}{n+1}+\frac{1}{n+2}-\frac{1}{2} \right)\)

\(\displaystyle \frac{1}{2}\left(\frac{1}{2}+\frac{1}{(n+1)(n+2)} \right)\)

\(\displaystyle \frac{1}{4}+\frac{1}{2(n+1)(n+2)}\)