TIFR 2014 Problem 5 Solution -Convergence of sequence

TIFR 2014 Problem 5 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India's premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
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Problem:

Let (a_n= (n+1)^{100}e^{-\sqrt{n}}) for (n \ge 1). Then the sequence (a_n) is

A. unbounded

B. bounded but not convergent

C. bounded and converges to 1

D. bounded and converges to 0

Discussion:

There can be various ways to do this. One way would be to check limit of

(f(x)=(x+1)^{100}e^{-\sqrt{x}}) as (x\to \infty) using L'Hospital rule.

One other way would be to use some inequalities involving (e^{-\sqrt{n}}) and ((n+1)^{100}).

We use the first approach.

( \lim_{x\to\infty}f(x)) is in (\infty /\infty) form.

We apply L'Hospital rule.

The derivative of (e^{\sqrt x} ) is (e^{\sqrt x}\frac{1}{2\sqrt x })

The derivative of ((x+1)^{100}) is (100(x+1)^{99}).

So ( \lim_{x\to\infty}f(x) = lim \frac{200(x+1)^{99}x^{1/2} }{e^{\sqrt x} } )

This is again in the (\infty / \infty ) form. So we continue this process.

Note that:

• At each step, the power of x in the numerator decreases by 1/2,
• and the \(e^{\sqrt x } \) remains as it is in the denominator.

Thus, after finitely many steps, we are left with an x with a power of 0 in the numerator. That is after finitely many steps, the numerator will become constant, while the denominator is still (e^{\sqrt x } ).

Thus the limit becomes 0.

So we conclude that the given sequence (a_n) converges to 0. Also, since any convergent sequence is bounded, this sequence is bounded. So option D is correct.

Helpdesk

• What is this topic: Real Analysis
• What are some of the associated concept: L'Hospital rule, Convergent Sequence, Bounded Sequence
• Book Suggestions: Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert