How are they getting this?

[kdinser]

This is in the series and convergence chapter

infinit sum (2^(n)+1)/2^(n+1)

lim as n goes to infinity of
(2^(n)+1)/2^(n+1) = \frac{1+2^{-n}}{2}=1/2

couldn't get latex to work right for the first part.

[himanshu121]

Lim_{n \inf} {(\frac{1}{2})}^n = 0

[arildno]

You have:
\frac{2^{n}+1}{2^{n+1}}=\frac{1}{2}\frac{2^{n}+1}{ 2^{n}}=\frac{1}{2}(1+2^{-n})

[kdinser]

Thanks, my algebra still seems to be a little rusty.

forgot that 2^{n+1}=2^n * 2^1

makes sense now and so do a few other ones that have been giving me headaches this morning.