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

I don't even know where to start.

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- Apr 11th 2010, 07:01 AMAmberosia32[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 AMArchie Meade
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 $\displaystyle \frac{1}{\left(\frac{2}{3}\right)^{|n|}}=\left(\fr ac{3}{2}\right)^{|n|}$ - Apr 11th 2010, 07:51 AMAmberosia32
- Apr 11th 2010, 08:01 AMArchie Meade
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 ? - Apr 11th 2010, 08:05 AMAmberosia32