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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