# Geometric Series

• Nov 4th 2007, 07:33 AM
slevvio
Geometric Series
I was having trouble with this question and wondered if I could get some help with it :)?

How many terms of the geometric series $1 + \frac{2}{3} + \frac{4}{9}$ + ... must be taken so that the sum differs from 3 by less than $10^{-3}$?

I started with this inequality:

$
3 - 10^{-3} < \frac{(\frac{2}{3})^n - 1}{\frac{2}{3} - 1} < 3 + 10^{-3}
$

and ended up with this:

$
\frac{ln(\frac{-1}{3000})}{ln(\frac{2}{3})} < n < \frac{ln(\frac{1}{3000})}{ln(\frac{2}{3})} = 19.74
$

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
• Nov 4th 2007, 10:21 AM
OnMyWayToBeAMathProffesor
codes
here is the page the explains codes.

3 - 10^(-3 < (2/3)^n - 1 )/(2/3 - 1) < 3 + 10 ^-3 would be

$3 - 10^ < \frac{(\frac{2}{3})^n - 1 )}{(\frac{2}{3} - 1) < 3 + 10^ (-3}$

u have to play with it.
• Nov 4th 2007, 08:33 PM
Soroban
Hello, slevvio!

I discovered that there is a strange step in this solution . . .

Quote:

How many terms of the geometric series $1 + \frac{2}{3} + \frac{4}{9} + \cdots$
must be taken so that the sum differs from 3 by less than $10^{-3}$ ?

The sum of the first n terms is: . $S_n \;=\;\frac{1 - (\frac{2}{3})^n}{1-\frac{2}{3}} \;=\;3\left[1-\left(\frac{2}{3}\right)^n\right]$

We want: . $\left|\,3\left[1-\left(\frac{2}{3}\right)^n\right] - 3\,\right| \;<\;0.001$

Then: . $-0.001 \;<\;3\left[1 - \left(\frac{2}{3}\right)^n\right] - 3 \;<\;0.001$

Divide by 3: . $-\frac{0.001}{3} \;<\;1 - \left(\frac{2}{3}\right)^n - 1 \;<\;\frac{0.001}{3}\quad\Rightarrow\quad -\frac{0.001}{3} \;<\:\underbrace{-\left(\frac{2}{3}\right)^n \;<\;\frac{0.001}{3}}$

The right half of the inequality can be discarded.
. . (Of course, a negative fraction is less than a positive fraction.)

So we have: . $-\frac{0.001}{3} \;<\;-\left(\frac{2}{3}\right)^n \quad\Rightarrow\quad
-\left(\frac{2}{3}\right)^n \;>\;-\frac{0.001}{3}$

Multiply by -1: . $\left(\frac{2}{3}\right)^n \;< \;\frac{0.001}{3}$

Take logs: . $\ln\left(\frac{2}{3}\right)^n \;<\;\ln\left(\frac{0.001}{3}\right)\quad\Rightarr ow\quad n\cdot\ln\left(\frac{2}{3}\right) \;<\;\ln\left(\frac{0.001}{3}\right)$

Divide both sides by $\ln\left(\frac{2}{3}\right)$ . . . note that this quantity is negative.

. . $n \;> \;\frac{\ln\left(\frac{0.001}{3}\right)}{\ln\left( \frac{2}{3}\right)} \;= \;19.7461...$

Therefore: . $n \:\geq\:20$

• Nov 5th 2007, 02:08 AM
slevvio
Thank you you certainly are a super member!