here is the page the explains codes.
3 - 10^(-3 < (2/3)^n - 1 )/(2/3 - 1) < 3 + 10 ^-3 would be
u have to play with it.
I was having trouble with this question and wondered if I could get some help with it ?
How many terms of the geometric series + ... must be taken so that the sum differs from 3 by less than ?
I started with this inequality:
and ended up with this:
Of course you cannot take the log of a negative number so how do I work out what n must be greater than? Also sorry I have no idea how to do that maths code thing so it might look a bit messy, Thanks
Hello, slevvio!
I discovered that there is a strange step in this solution . . .
How many terms of the geometric series
must be taken so that the sum differs from 3 by less than ?
The sum of the first n terms is: .
We want: .
Then: .
Divide by 3: .
The right half of the inequality can be discarded.
. . (Of course, a negative fraction is less than a positive fraction.)
So we have: .
Multiply by -1: .
Take logs: .
Divide both sides by . . . note that this quantity is negative.
. .
Therefore: .