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

I don't even know where to start.

Printable View

- Apr 11th 2010, 08:01 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, 08:18 AMArchie Meade
It depends on n.

If n starts from zero, then

This is the largest value that 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

increase to infinity as n goes to minus infinity.

This is because - Apr 11th 2010, 08:51 AMAmberosia32
- Apr 11th 2010, 09:01 AMArchie Meade
Hi Amberosia32,

you do need to clarify your question however.

My answer only takes into account the graph of

This may not be what you are looking for,

as your post mentions "geometric series".

If so, and is__not__the formula for the sum,

ie it may be a series of the form

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, 09:05 AMAmberosia32