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 $\forall \epsilon>0:0+\epsilon>a_{n}(\epsilon)$. How do I go about finding $a_{n}(\epsilon)$?
 A: 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 $0<\frac{1}{n+10}$. $\therefore$ $0$ is the lower bound. Next, we need to prove $0$ is the greatest lower bound. Suppose $\epsilon>0$ be another lower bound of the set. $$\epsilon<\frac{1}{n+10}$$
$$\implies n<\frac{1}{\epsilon}-10$$. 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.
A: Lower bound is $0$.
Show that $\inf( A) =0$, I.e. $0$ is the greatest lower bound.
Proof:
Assume $\epsilon \gt 0$, $\epsilon$ real, is a lower bound.
Archimedes' Axiom:
There exists a positive integer (natural number) such that
$n_0 + 1 \gt 1/\epsilon.$
Hence:
For $n\ge n_0:$
$\dfrac{1}{n+1} \le \dfrac{1}{n_0+1} \lt \epsilon$, 
Contradiction.
A: Clearly $0$ is a lower bound. Given $\varepsilon>0$, observe that if $n>1/\varepsilon $
$$
\frac{1}{n+10}\leq \frac{1}{n}<\epsilon.
$$
Hence $\varepsilon$ is not a lower bound and $0$ is the greatest lower bound i.e. the infimum.
A: Start here $${1\over n + 10} < \epsilon.$$
Take reciprocals; note that this reverses order
$$n + 10 > {1\over \epsilon}$$
