Given (an)n≥1 a sequence of real numbers, which of the following statements is true?

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CSIR-UGC (NET) Mathematical Science: Held on (26 Nov 2020)
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  1. \(\sum_{n ≥ 1}(-1)^n \frac{a_n}{1+\left|a_n\right|}\) converges 
  2. There is a subsequence \(\left(a_{n_k}\right)_{k \geq 1}\) such that \(\sum_{k ≥ 1} \frac{a_{n_k}}{1+\left|a_{n_k}\right|}\) converges
  3. There is a number b such that \(\sum_{n \geq 1}\left|b-\frac{a_n}{1+\left|a_n\right|}\right|(-1)^n\) converges
  4. There is a number b and a subsequence \(\left(a_{n_k}\right)_{k \geq 1}\) such that \(\sum_{k \geq 1}\left|b-\frac{a_{n_k}}{1+\left|a_{n_k}\right|}\right|\) converges

Answer (Detailed Solution Below)

Option 4 : There is a number b and a subsequence \(\left(a_{n_k}\right)_{k \geq 1}\) such that \(\sum_{k \geq 1}\left|b-\frac{a_{n_k}}{1+\left|a_{n_k}\right|}\right|\) converges
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Explanation:

Tips: Try to discard options by taking suitable choice for < an >.

option (1). Let an = 1 then \(\operatorname{\lim}_{n \rightarrow \infty}\)(-1)n .\(\frac{1}{2}\) ≠ 0

option (2). Let an = <1> and \(a_{n_k}\) = <1> then

\(\sum_{k \geq 1}^{\infty} \frac{a_{n_k}}{1+\left|a_{n_k}\right|}=\sum_{k=1}^{\infty} \frac{1}{1+1}=\sum \frac{1}{2}\)  Not convergent.

option (3), option (4): (NB: may be you have to try with more) then one sea, <an>

Let an = (-1)n then \(\sum\left|b-\frac{a_n}{1+| a_n \mid}\right|(-1)^n\)

\(=\sum\left|b-\frac{(-1)^n}{2}\right|(-1)^n\)

But here fixed 'b S.t above series become cgt. You may take b = ½ or = -½ but not both otherwise uniqueness will be lost.

⇒ option (3) is false. 

option (4): As discussed earlier. take b = ½ and \(a_{n_k}\) = <1> then about series becomes convergent. Hence option (4) is true.

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