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Summation of series

Erfan

New member
Jul 19, 2013
9
Prove that 1/(1*2*3) + 1/(2*3*4) + ... + 1/(n*(n+1)*(n+2)) = 1/4 - 1/(2*(n+1)*(n+2)
.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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?
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
Prove that 1/(1*2*3) + 1/(2*3*4) + ... + 1/(n*(n+1)*(n+2)) = 1/4 - 1/(2*(n+1)*(n+2)
.
This looks like a telescopic series to me...
 

Erfan

New member
Jul 19, 2013
9
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?
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 :D
 

chisigma

Well-known member
Feb 13, 2012
1,704
Prove that 1/(1*2*3) + 1/(2*3*4) + ... + 1/(n*(n+1)*(n+2)) = 1/4 - 1/(2*(n+1)*(n+2)
.
It is convenient to proceed by induction. Let's suppose that...


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

... 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


$\chi$ $\sigma$
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,708
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 :D
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 $0$.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I would write the sum as:

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

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:

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

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

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