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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?

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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 $n \geq n_0$.

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  • $\begingroup$ Thanks a lot, FYI, its $\frac{11}{14}$. I made a calculation error before. $\endgroup$ – Umesh shankar Jun 21 at 9:22
  • $\begingroup$ Also is this justification required every time? Can we say if a monotone decreasing sequence is convergent to $L$, then $ inf(A)=L$, since it is well known that bounded monotone sequences converge to either Supremum or infimum. $\endgroup$ – Umesh shankar Jun 21 at 9:26
  • $\begingroup$ @Umeshshankar That should be enough, but it all depends on your teacher. $\endgroup$ – Kavi Rama Murthy Jun 21 at 9:27
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You just took the opposite value of $\epsilon$. It should be $0<\epsilon<m$. Take this value and get the answer

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