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first I pluged in n values and found that as n gets bigger x(sub n) gets smaller and appears to go to zero. so I know I want to show/prove that the sequence is monotone decreasing and is bounded by zero. by induction i find that when n=1 , x(sub n)= 1/2. and i have x(sub n+1)= (n+1)/2^(n+1).

(this is about where i start to feel less confident, I'm not sure if i'm using induction properly and also on my next step i'm stuck on manipulating the inequality).

now I have x(sub n+1)

__<__x(sub n). substituting and rearanging the inequality I get:

[(n+1)/2^(n+1)]-[n/2^n]

__<__0

[(2^n)(n+1)-(n)(2^(n+1)]/(2^n)(2^(n+1))

__<__0

This is where I am stuck, my strong points do not lie with manipulating inequalities with variables in the exponets (any links or suggestions to help me view examples of tricks of the trade here are welcome) also is this sufficient enough to state that it has a lower bound and the lim n/2^n = 0 as n goes to infinity?