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

- 1
- 0
- \(+\infty\)
- \(1/2\)

TIFR 2019 GS Part A, Problem 6

Limit and Integration

Hard

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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