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.
The image is a front cover of a book named Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert. This book is very useful for the preparation of TIFR Entrance.
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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
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:
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.
