a(n) = (2^(n-1)) / ((2^n)-1) with n --> infinite. thank you
Last edited by nappysnake; Nov 27th 2011 at 12:24 PM.
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Originally Posted by nappysnake a(n) = (2^(n-1)) / ((2^n)-1) with n --> infinite. $\displaystyle \lim_{n \to \infty} \frac{2^{n-1}}{2^n - 1} =$ divide every term by $\displaystyle 2^{n-1}$ ... $\displaystyle \lim_{n \to \infty} \frac{1}{2 - \frac{1}{2^{n-1}}} = \frac{1}{2}$
thank you skeeter for letting me come to a crucial realization, it was very clear
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