# Understand the problem

The limit $$\lim_{n \to \infty}n^2 \int_0^1\frac{1}{(1+x^2)^n}$$ is equal to
• 1
• 0
• $$+\infty$$
• $$1/2$$

##### Source of the problem
TIFR 2019 GS Part A, Problem 6

##### Topic
Limit and Integration

Hard

### Introduction to Real Analysis by Donald R. Sherbert and Robert G. Bartle

Do you really need a hint? Try it first!

Can you try with beta function? Let me give you some formulas: $$\int_0^{\pi/2}\cos^{2n-2}\theta d\theta=\frac 12B(\frac 12(0+1), \frac 12(2n-2+1)=\frac 12B(\frac 12, \frac {2n-1}{2})$$$$=\frac 12\frac{\Gamma({\frac 12})\Gamma(\frac {2n-1}{2})}{\Gamma(n+1)}=\frac 12\sqrt{\pi}(\frac{2n-1}{2}-1)\frac{\Gamma(\frac {2n-1}{2}-1)}{n!}$$
$$\frac 12\sqrt{\pi}(\frac{2n-1}{2}-1)\frac{\Gamma(\frac {2n-1}{2}-1)}{n!}=\frac 12\sqrt{\pi}\frac{2n-3}2\times \frac{2n-5}2\times \cdots \frac{1}2\times \sqrt{\pi}$$ $$=\frac{\pi}2 \times \frac{(2n-2)(2n-3)\cdots 2.1}{(2n-2)(2n-4)\cdots 2 \times 2^{n-1}}$$

$$\frac{\pi}2 \times \frac{(2n-2)(2n-3)\cdots 2.1}{(2n-2)(2n-4)\cdots 2 \times 2^{n-1}}$$

$$=\frac{\pi}{2.4^{n-1}} \times \frac{n}{2n} \times \frac{(2n)!}{n!(n-1)!} \times \frac{1}{2n-1}=\frac{n\pi}{(2n-1).4^{n}} \times {2n \choose n}$$

Now $$\frac{n\pi}{(2n-1).4^{n}} \times {2n \choose n}=\frac{1}{4^{n}} \times {2n \choose n}\sim \frac{1}{\sqrt{\pi n}}$$, so $$n^2\frac{1}{\sqrt{\pi n}} \to \infty$$

Put $$x=\tan(\theta)$$ in the main expression. And observe how much less the expression would be from $$\int_0^{\pi/2}\cos^{2n-2}\theta d\theta$$.

$$\int_0^{\pi/2}\cos^{2n-2}\theta d\theta=\int_0^{\pi/4}\cos^{2n-2}\theta d\theta+\int_{\pi/4}^{\pi/2}\cos^{2n-2}\theta d\theta<\int_0^{\pi/4}\cos^{2n-2}\theta d\theta+\frac{1}{2^{2n-2}}\frac{\pi}{4}$$

#### $$\int_{0}^{\pi/4}\cos^{2n-2}\theta d\theta$$ differs by a finite no from $$\int_0^{\pi/2}\cos^{2n-2}\theta d\theta$$ hence the limit will still be infinity.

So we are done with the completion of Hint 5 and as well as the solution of the question

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