Question

How do I find the greatest lower bound for the sequence `A={\frac{1}{n+10}}_{n=1}^{\infty}`?

I know the infimum is zero, and I know I need to find `a_{n}(\epsilon)\in A` such that `\forall \epsilon>0:0+\epsilon>a_{n}(\epsilon)`. How do I go about finding `a_{n}(\epsilon)`?

Answer 1

supremum `=1/11` and infimum `= 0`

Explanation:

We seek upper and lower bounds for the sequence:

`{ 1/(n+10) }` for `n in [1,oo)`

If we consider the function:

`f(x) = x+10`

Then trivially, `f(x)` is monotone increasing and so `1/f(x)` is monotone decreasing.

Thus we have an upper bound when `n=1` corresponding to:

supremum `=1/f(1)=1/11`

And a lower bound as `n rarr oo` corresponding to:

infimum `=lim_(n rarr oo) 1/f(n) = lim_(n rarr oo) 1/(n+10) = 0`