Determining convergence, monoticity and possible bounds

[Nerpilis]

my problem is to determine if a sequence is convergent by deciding on wether its monotone increasing or decreasing, and to identify a bound if it exists. given x(sub n) = n/2^n.

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?

 

[Muzza]

Do you know that 2^(n + 1) = 2 * 2^n? Hence

(n + 1)/2^(n + 1) < n/2^n
<=>
(n + 1)/(2 * 2^n) < n/2^n

and you can cancel 1/2^n from both sides, leaving something which is very easy to prove.

 

[Nerpilis]

wow ok that is very clever, it was staring at me the whole time and i missed it. now i have the inequality to show:
(1-n)/(2)(2^n)<0<n/2^n for n>1 then that would definitely make x(sub n+1)<x(sub n) thus proving monotone decreasing convergence. I know that this sequence's limit is zero and is bounded by zero but I' not sure how to 'prove' it besides just looking at it and saying it goes to zero.

 

[Muzza]

If you know the binomial theorem, you can use that to estimate 2^n (i.e (1 + 1)^n), and then use the squeeze theorem...

But a more elementary approach is to find a recursive formula relating x_(n + 1) to x_n. If you take limits on both sides of the recurrence equation, everything should fall out nicely.