# [SOLVED] Geometric series and bounding

• Apr 11th 2010, 07:01 AM
Amberosia32
[SOLVED] Geometric series and bounding
Find the values that bound (above and below) the geometric series (2/3)^n

I don't even know where to start.
• Apr 11th 2010, 07:18 AM
Quote:

Originally Posted by Amberosia32
Find the values that bound (above and below) the geometric series (2/3)^n

I don't even know where to start.

It depends on n.
If n starts from zero, then $\left(\frac{2}{3}\right)^0=1$

This is the largest value that $\left(\frac{2}{3}\right)^n$ can be,
since to form higher powers we are multiplying by a value <1 continuously.
As n approaches infinity, the expression approaches zero.

You should have an idea of what the graph of the curve looks like,
as n ranges from 0 to infinity.

If n is allowed to be negative, then

$\left(\frac{2}{3}\right)^{n}$ increase to infinity as n goes to minus infinity.

This is because $\frac{1}{\left(\frac{2}{3}\right)^{|n|}}=\left(\fr ac{3}{2}\right)^{|n|}$
• Apr 11th 2010, 07:51 AM
Amberosia32
Quote:

It depends on n.
If n starts from zero, then $\left(\frac{2}{3}\right)^0=1$

This is the largest value that $\left(\frac{2}{3}\right)^n$ can be,
since to form higher powers we are multiplying by a value <1 continuously.
As n approaches infinity, the expression approaches zero.

You should have an idea of what the graph of the curve looks like,
as n ranges from 0 to infinity.

If n is allowed to be negative, then

$\left(\frac{2}{3}\right)^{n}$ increase to infinity as n goes to minus infinity.

This is because $\frac{1}{\left(\frac{2}{3}\right)^{|n|}}=\left(\fr ac{3}{2}\right)^{|n|}$

ok, thanks
• Apr 11th 2010, 08:01 AM
Hi Amberosia32,

you do need to clarify your question however.
My answer only takes into account the graph of $f(n)=\left(\frac{2}{3}\right)^n$

This may not be what you are looking for,
as your post mentions "geometric series".

If so, and $\left(\frac{2}{3}\right)^n$ is not the formula for the sum,

ie it may be a series of the form $a+a\frac{2}{3}+a\left(\frac{2}{3}\right)^2+......$

If the series is something like this, then it needs to be evaluated first.

Hence, do you have an entire question on this ?
• Apr 11th 2010, 08:05 AM
Amberosia32
Quote:

Hi Amberosia32,

you do need to clarify your question however.
My answer only takes into account the graph of $f(n)=\left(\frac{2}{3}\right)^n$

This may not be what you are looking for,
as your post mentions "geometric series".

If so, and $\left(\frac{2}{3}\right)^n$ is not the formula for the sum,

ie it may be a series of the form $a+a\frac{2}{3}+a\left(\frac{2}{3}\right)^2+......$

If the series is something like this, then it needs to be evaluated first.

Hence, do you have an entire question on this ?

I meant sequence. I'm sorry. And n start at 1, so it would be bounded between 2/3 and 0. I hope. lol.