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Summation of series
- Thread starter Erfan
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- #2
This looks like a telescopic series to me...
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- #4
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
chisigma
Well-known member
- Feb 13, 2012
- 1,704
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
$\chi$ $\sigma$
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- #6
- Feb 7, 2012
- 2,784
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
- Admin
- #7
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:
\(\displaystyle \frac{1}{2}\left(1-\frac{1}{n+1}+\frac{1}{n+2}-\frac{1}{2} \right)\)
\(\displaystyle \frac{1}{2}\left(\frac{1}{2}+\frac{1}{(n+1)(n+2)} \right)\)
\(\displaystyle \frac{1}{4}+\frac{1}{2(n+1)(n+2)}\)
\(\displaystyle \sum_{k=1}^n\left(\frac{1}{k(k+1)(k+2)} \right)\)
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:
\(\displaystyle \frac{1}{2}\left(1-\frac{1}{n+1}+\frac{1}{n+2}-\frac{1}{2} \right)\)
\(\displaystyle \frac{1}{2}\left(\frac{1}{2}+\frac{1}{(n+1)(n+2)} \right)\)
\(\displaystyle \frac{1}{4}+\frac{1}{2(n+1)(n+2)}\)