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About LinkBacksFind 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.Originally Posted by Amberosia32
If n starts from zero, then
This is the largest value thatcan 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^{|n|}}=\left(\fr ac{3}{2}\right)^{|n|})
ok, thanksIt depends on n.
If n starts from zero, then
This is the largest value that
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
This is because
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
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 ?
I meant sequence. I'm sorry. And n start at 1, so it would be bounded between 2/3 and 0. I hope. lol.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
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 ?
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