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IIT JAM MS 2020 Section A Problem 1 Solution


If \{x_{n}\}_{n \geq 1} is a sequence of real numbers such that \lim _{n \rightarrow \infty} \frac{x_{n}}{n}=0.001, then

(A) \{x_{n}\}_{n \geq 1} is a bounded sequence
(B)\{x_{n}\}_{n \geq 1} is an unbounded sequence
(C) \{x_{n}\}_{n \geq 1} is a convergent sequence
(D) \{x_{n}\}_{n \geq 1} is a monotonically decreasing sequence


Hint 1

If \{x_{n}\}_{n \geq 1} was bounded, show that \lim _{n \rightarrow \infty} \frac{x_{n}}{n}=0, by sandwich theorem.

Hint 2

If \{x_{n}\}_{n \geq 1} was convergent, show that \lim _{n \rightarrow \infty} \frac{x_{n}}{n}=0, by algebra of limits.

Hint 3

If \{x_{n}\}_{n \geq 1} was motonotically decreasing and bounded below, then it would have been convergent by Monotone Convergence Theorem.

Let's consider if it is not below below, i.e. \lim _{n \rightarrow \infty} {x_{n}} = -\infty

  • Take {x_{n}} = -n
  • Take {x_{n}} = -n^2
  • Take {x_{n}} = - \sqrt{n}

Find the limit in each of this case.

Hence, it will be unbounded. See the full solution and proof idea below.

Full Solution

Food For Thoughts

  • What if, \lim _{n \rightarrow \infty} \frac{x_{n}}{n}= 0?
  • Find examples.

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