How are they getting this?

1. Nov 6, 2004

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.

Last edited: Nov 6, 2004
2. Nov 6, 2004

himanshu121

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

Last edited: Nov 6, 2004
3. Nov 6, 2004

arildno

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

Last edited: Nov 6, 2004
4. Nov 6, 2004

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.