How to find the infimum of $A=\{\frac{1}{n+10}\}_{n\in\mathbb N}$

Question

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

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.

Answer 2

Start here $${1\over n + 10} < \epsilon.$$ Take reciprocals; note that this reverses order $$n + 10 > {1\over \epsilon}$$

Answer 3

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.

Answer 4

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.