How to find the infimum of $A=\{\frac{1}{n+10}\}_{n\in\mathbb N}$ I know the infimum is zero, and I know I need to find $a_{n}(\epsilon)\in A$ such that $\forall \epsilon&gt;0:0+\epsilon&gt;a_{n}(\epsilon)$. How do I go about finding $a_{n}(\epsilon)$?
Method 1:-

$\because$ Given sequence is monotonically decreasing, so, the sequence $\{\frac{1}{n+10}\}$ converges to its infimum.

Method 2:-
Infimum is the greatest lower bound. We know that $0&lt;\frac{1}{n+10}$. $\therefore$ $0$ is the lower bound. Next, we need to prove $0$ is the greatest lower bound. Suppose $\epsilon&gt;0$ be another lower bound of the set. $$\epsilon&lt;\frac{1}{n+10}$$
$$\implies n&lt;\frac{1}{\epsilon}-10$$. It is true for all Natural numbers. natural numbers are not bounded. So, $\epsilon$ cannot be the lower bound. $\epsilon$ was an arbitrary positive real number.
Start here $${1\over n + 10} &lt; \epsilon.$$
Take reciprocals; note that this reverses order
$$n + 10 &gt; {1\over \epsilon}$$