# 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 ${a}_{n} \left(\setminus \epsilon\right) \setminus \in A$ such that $\setminus \forall \setminus \epsilon > 0 : 0 + \setminus \epsilon > {a}_{n} \left(\setminus \epsilon\right)$. How do I go about finding ${a}_{n} \left(\setminus \epsilon\right)$?

May 2, 2018

supremum $= \frac{1}{11}$ and infimum $= 0$

#### Explanation:

We seek upper and lower bounds for the sequence:

$\left\{\frac{1}{n + 10}\right\}$ for $n \in \left[1 , \infty\right)$

If we consider the function:

$f \left(x\right) = x + 10$

Then trivially, $f \left(x\right)$ is monotone increasing and so $\frac{1}{f} \left(x\right)$ is monotone decreasing.

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

supremum $= \frac{1}{f} \left(1\right) = \frac{1}{11}$

And a lower bound as $n \rightarrow \infty$ corresponding to:

infimum $= {\lim}_{n \rightarrow \infty} \frac{1}{f} \left(n\right) = {\lim}_{n \rightarrow \infty} \frac{1}{n + 10} = 0$