# Infimum of a sequence proof by contradiction

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

Any help from here?

You made a wrong choice for $$\epsilon$$. Take $$0 <\epsilon . Then you get the contradiction $$m =\inf A \leq a_n <\frac {11} {14} +\epsilon when $$n \geq n_0$$.
• Thanks a lot, FYI, its $\frac{11}{14}$. I made a calculation error before. – Umesh shankar Jun 21 at 9:22
• 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. – Umesh shankar Jun 21 at 9:26
You just took the opposite value of $$\epsilon$$. It should be $$0<\epsilon. Take this value and get the answer