Find the supremum and infimum of the following set

Let A = {n^2/2^n | n in N}. Find the supremum and infimum, proving your assertions.

**Attempt at Solution**

SupA:

The highest term in the set is 9/8 and thus it is the supremum, by definition.

InfA:

The terms in the set approach 0 as n becomes arbitrarily large. It is clear that 0 < n^2/2^n for all n, thus, 0 is a lower bound.

I'm not sure how to prove that 0 is the infimum. Can someone lead me in the right direction?

Re: Find the supremum and infimum of the following set

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**My Little Pony** Let A = {n^2/2^n | n in N}. Find the supremum and infimum, proving your assertions.

InfA:

The terms in the set approach 0 as n becomes arbitrarily large. It is clear that 0 < n^2/2^n for all n, thus, 0 is a lower bound.

I'm not sure how to prove that 0 is the infimum. Can someone lead me in the right direction?

You have already shown that 0 is a lower bound for the set.

Now show if $\displaystyle c>0$ then there is a $\displaystyle k\in\mathbb{N}$ such that $\displaystyle 0<\frac{k^2}{2^k}<c$.

Thus proving that no number greater that 0 is a lower bound.

That can be done by using the fact that $\displaystyle \left(\frac{n^2}{2^n}\right)\to 0.$