partial sums of sequences

**confusedgirl** · March 2nd 2010, 06:18 AM

find the Sn of An = (1/n) -(1/n+2)
i got the denominator n^2 + 3n + 2 but can't find the numerator

**Krizalid** · March 2nd 2010, 06:52 AM

$\frac{1}{n}-\frac{1}{n+2}=\left( \frac{1}{n}-\frac{1}{n+1} \right)+\left( \frac{1}{n+1}-\frac{1}{n+2} \right),$ so $\sum\limits_{n=1}^{m}{\left( \frac{1}{n}-\frac{1}{n+2} \right)}=\sum\limits_{n=1}^{m}{\left( \frac{1}{n}-\frac{1}{n+1} \right)}+\sum\limits_{n=1}^{m}{\left( \frac{1}{n+1}-\frac{1}{n+2} \right)}.$

can you finish?

**confusedgirl** · March 2nd 2010, 07:01 AM

sorry, but no.

**Krizalid** · March 2nd 2010, 10:10 AM

okay then, try to look at what a telescoping sum is.

**Archie Meade** · March 2nd 2010, 11:24 AM

Originally Posted by confusedgirl

find the Sn of An = (1/n) -(1/n+2)
i got the denominator n^2 + 3n + 2 but can't find the numerator

Hi confusedgirl,

You can write it this way...by placing in the values of n=1, 2, 3, 4, 5...

$A_n=\frac{1}{n}-\frac{1}{n+2}$

$S_n=A_1+A_2+A_3+A_4+A_5+....A_{n-2}+A_{n-1}+A_n$

$S_n=\left(\frac{1}{1}-\frac{1}{1+2}\right)+\left(\frac{1}{2}-\frac{1}{2+2}\right)+\left(\frac{1}{3}-\frac{1}{3+2}\right)+\left(\frac{1}{4}-\frac{1}{4+2}\right)......$

$=\left(1-\frac{1}{3}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{4}-\frac{1}{6}\right)+$ $.....+\left(\frac{1}{n-2}-\frac{1}{n}\right)+\left(\frac{1}{n-1}-\frac{1}{n+1}\right)+\left(\frac{1}{n}-\frac{1}{n+2}\right)$

If you examine this, you will see that virtually all of the terms cancel,
due to the same terms having + and -.

Which terms will be left standing?

Also, what process did you use to calculate your original denominator?

**confusedgirl** · March 2nd 2010, 03:04 PM

i'm sorry, but i think my question wasn't clear i need to get Sn where
Sn = Sn-1 + An
like
S1 = 1-1/3 = 2/3
S2 = 2/3 +1/2 -1/4 = 11/12
S3= 21/20
S4 = 34/30

i really didn't use any process or technique to get the denominator but just figured it out by trial and error

**confusedgirl** · March 2nd 2010, 03:05 PM

Originally Posted by Krizalid

okay then, try to look at what a telescoping sum is.

ok

**Archie Meade** · March 2nd 2010, 03:38 PM

$1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+\frac{1}{4}-\frac{1}{6}+...\frac{1}{n-2}-\frac{1}{n}+\frac{1}{n-1}-\frac{1}{n+1}+\frac{1}{n}-\frac{1}{n+2}$

reduces to

$1+\frac{1}{2}-\frac{1}{n+1}-\frac{1}{n+2}=\frac{3}{2}-\left(\frac{1}{n+1}+\frac{1}{n+2}\right)$

since the last term to cancel is $\frac{1}{n}$

as a term can only cancel a term 3 places to the right.

$=\frac{3}{2}-\frac{n+2+(n+1)}{(n+1)(n+2)}=\frac{3}{2}-\frac{2n+3}{(n^2+3n+2)}$

as all the other terms cancel.

This will give you Sn for any n, without having to calculate a preceding sum.

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