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Suppose n=3
then 3^2=9>2*3=6
Now let n=4
4^2=16>8
Now let’s look at n+1
(n+1)^2=n^2+2n+1
n^2+2n+1>2(n+1)
n^2>1 which is true for every n greater than 2 or 2. As we looked at n+1: 2+1=3 thus it is true for n>=3.
QED
Also if n is real, not natural there is another way to proove this
Let’s look at the derivatives of f(n)=n^2 and g(n)=2n (functions are continuous R->R)
f’(n)=2n; g’(n)=2
Values of them are equal only in one point if n=1.
If n>1 than monotonicaly increasing 2n is bigger than constant function 2.
The original functions intersect in
n^2=2n
n=0; n=2
Functions f and g have the same values in n=2 but f’ is bigger. That means they start from one point, but n^2 always increases faster for n>2. Thus, it will have bigger values and the gap will be increasing. Lim(n^2/2n) n->inf diverges.
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