• No products in the cart.

# 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.

October 9, 2017