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 $\displaystyle \left(\frac{2}{3}\right)^0=1$
This is the largest value that $\displaystyle \left(\frac{2}{3}\right)^n$ 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
$\displaystyle \left(\frac{2}{3}\right)^{n}$ increase to infinity as n goes to minus infinity.
This is because $\displaystyle \frac{1}{\left(\frac{2}{3}\right)^{|n|}}=\left(\fr ac{3}{2}\right)^{|n|}$
Hi Amberosia32,
you do need to clarify your question however.
My answer only takes into account the graph of $\displaystyle f(n)=\left(\frac{2}{3}\right)^n$
This may not be what you are looking for,
as your post mentions "geometric series".
If so, and $\displaystyle \left(\frac{2}{3}\right)^n$ is not the formula for the sum,
ie it may be a series of the form $\displaystyle a+a\frac{2}{3}+a\left(\frac{2}{3}\right)^2+......$
If the series is something like this, then it needs to be evaluated first.
Hence, do you have an entire question on this ?