Sequence Question Real analysis IIT JAM 2018

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General Mathematics


Question: 

Solution:

Note that,

  •  An element $m$ is maximum of set $S$ iff $s≤m$ for all $s\in S$ and $m\in S$. Further, if a set $S$ has maximum element then it is supremum of $S$.
  • Similarly, an element $l$ is minimum or least element of set $S$ if $l≤s$ all $s\in S$ and $l\in S$. Further, if a set $S$ has minimum or least element then it is infimum of $S$.

Hence,

$sup\{a_n: n\in\mathbb{N}\}$
$=sup\{a_1,a_2,a_3,...\}$
$=sup\{3,\frac{5}{4},\frac{5}{3},\frac{17}{16},...\}$
$=3$ (since maximum  of set $\{a_1,a_2,...\}$ is 3.   )

Similarly,

 $inf\{a_n: n\in\mathbb{N}\}$
$=inf\{a_1,a_2,a_3,...\}$
$=inf\{3,\frac{5}{4},\frac{5}{3},\frac{17}{16},...\}$
$=1$ (since, $\{a_1,a_2,...\}$ has minimum or least element 1) 

Hence option (A) is true, (C) is false. 

Now to discuss options (B) and (C): we note that, 

$\text{lim inf} (a_n)$ is infimum of set of all subsequential limits of sequence $(a_n)$ and similarly, $\text{lim sup} (a_n)$ is supremum of set of all subsequential limits of sequence $(a_n)$

Hence clearly, $\text{lim inf}(a_n)=inf\{1,2\}=1$ and $\text{lim sup}(a_n)=sup\{1,2\}=2$ 
Hence both options (B), (D) are false! 

Hence answer is, option (A) only. 

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