Can you write the recursion formula for this series? $\displaystyle a_n=\frac{1}{2^n}$
Last edited by CaptainBlack; Jan 11th 2006 at 01:00 PM.
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You give: $\displaystyle a_n=\frac{1}{2^n}$. This as it stands is a sequence not a series, and $\displaystyle a_{n+1}=\frac{1}{2^{n+1}}=\frac{1}{2^n.2}=\frac{1} {2}a_n $. RonL
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