# Infinite diverging series

• July 2nd 2013, 05:57 PM
minneola24
Infinite 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
• July 2nd 2013, 06:13 PM
Prove It
Re: Infinite diverging series
\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 \begin{align*} \frac{1}{2} \end{align*} will eventually lead to \displaystyle \begin{align*} \infty \end{align*}.