# Thread: partial sums of sequences

1. ## partial sums of sequences

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

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

can you finish?

3. sorry, but no.

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

5. 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...

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

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

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

$\displaystyle =\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)+$$\displaystyle .....+\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?

6. 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

7. Originally Posted by Krizalid
okay then, try to look at what a telescoping sum is.
ok

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

reduces to

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

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

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

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

as all the other terms cancel.

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