Pre Calc
posted by Masha .
Write as a series and express as a rational number:
1. 5.36363636....
2. 0.0123123....
Use this series and find S1,S2,S3,S4,Sn, and lim Sn.
1/1*3 + 1/3*5 + 1/5*7+...+ 1/(2n1)(2n+1)

It works easier to have one type problem per question, as you will see here.
On the first, try something like multiping by a power of ten that will align digits. For instance, in the first, multiply by 100
536.36363636 . Subtract the original number
536.36365.363636= 531
Now divide by 99 (think out why)
531/99 is your fraction 
I would split it this way:
5.36363636....
= 5 + (.36 + .0036 + .000036 + ...}
so for the bracket part, a=.36, r = .01
remember S_{∞} = a/(1r)
= .36/(1.01) = .36/.99 = 36/99 = 4/11
then 5.36363636.... = 5 4/11 or 59/11
do the rest the same way 
1. 5.36363636.... = 5 + 0.36(1 + 10^2 + 10^4 + ..)
= 5 + 0.36[1/(1.01)]
= 5 + 0.36 * 100/99
= 5 + 36/99
2. Do it the same way
3. If you are dealing with sums, the first term is
S1 = 1/3
and the second term is
S2 = 1/3 + 1/15 = 6/15
Do the others and see what Sn and the limit are 
for your last question...
1/1*3 + 1/3*5 + 1/5*7+...+ 1/(2n1)(2n+1)
S1 = 1/3
S2 = 1/3 + 1/15 = 6/15 = 2/5
S3 = 2/5 + 1/35 = 15/35 = 3/7
do you see a pattern?
so what is Sn ?
An interesting question now would be,
Prove that your answer to the above is correct by induction.