Respuesta :

Ozgr
First note that [tex] \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}}[/tex]

If you take limit, then you have [tex] \lim_{n \to \infty}( \frac{1}{2} + \frac{1}{2^{n+1}})= \lim_{n \to \infty}( \frac{1}{2}) +\lim_{n \to \infty}(\frac{1}{2^{n+1}})=\frac{1}{2} +0= \frac{1}{2} [/tex]



Hi steve ;) 
 
you just have to apply simple exponent rule:
 [tex] \frac{x^n}{x^y} =x^{n-m} [/tex]

& RIP OS ;-; :( 
#os<3