Consider the sequence defined as $a_1=1$ and $$a_{n+1}=\frac{7a_n+11}{21}$$
It is evident that $\lim_{n \to \infty}a_n =\frac{11}{14}$ and its a monotone decreasing sequence.
Now can we say by Monotone convergence theorem, that the infimum of the set $A=\left\{a_n: n \in N\right\}$ as $\frac{11}{14}$ or should we justify that $\frac{11}{14}$ is the inf(A)?
Nevertheless i tried to justify by contracdiction. Let $inf(A)=m >\frac{11}{14}$.Since the sequence $\left\{a_n\right\}$ is convergent, we have for every $\epsilon >0$, $\exists$ $n_0\in N$ such that $\forall n \geq n_0$
$$\left|a_n-\frac{11}{14}\right|<\epsilon$$
Now choose $\epsilon \in \left(\frac{11}{14},m\right)$, So we have
$$\left|a_n-\frac{11}{14}\right|<\epsilon<m$$
Any help from here?
You made a wrong choice for $\epsilon$. Take $0 <\epsilon <m-\frac {11} {14}$. Then you get the contradiction $m =\inf A \leq a_n <\frac {11} {14} +\epsilon <m$ when $n \geq n_0$.