Having an issue with #17

Do I divide the top and bottom by 1/2^n

If so how does the bottom work since the exponet is n+1?Attachment 28715

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- Jul 2nd 2013, 05:57 PMminneola24Infinite diverging series
Having an issue with #17

Do I divide the top and bottom by 1/2^n

If so how does the bottom work since the exponet is n+1?Attachment 28715 - Jul 2nd 2013, 06:13 PMProve ItRe: Infinite diverging series
$\displaystyle \displaystyle \begin{align*} \frac{2^n + 1}{2^{n+1}} &= \frac{2^n}{2^{n+1}} + \frac{1}{2^{n+1}} \\ &= \frac{1}{2} + \frac{1}{2^{n+1}} \\ &= \frac{1}{2} + \left( \frac{1}{2} \right) ^{n+1} \end{align*}$

So the series must be divergent because constantly adding $\displaystyle \displaystyle \begin{align*} \frac{1}{2} \end{align*}$ will eventually lead to $\displaystyle \displaystyle \begin{align*} \infty \end{align*}$.