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Convergence of sequence (TIFR 2014 problem 5)

Question:

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.

 

 

 

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