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if the limit $$L=\lim_{n\to \infty}\sqrt n \int_0^1 \frac{1}{(1+x^2)^n}dx$$exists and is larger than $\frac{1}{2}$ then prove that $\frac{1}{2} < L < 2 $ .


marked as duplicate by StubbornAtom, Lee David Chung Lin, Lord Shark the Unknown, Jyrki Lahtonen, Cesareo Apr 14 at 8:58

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Using $(1+x^2)^n>1+nx^2$ for $x\geq 0$

So $$\frac{1}{(1+x^2)^n}<\frac{1}{1+nx^2}$$

So $$\lim_{n\rightarrow \infty}\sqrt{n}\int^{1}_{0}<\lim_{n\rightarrow \infty}\sqrt{n}\int^{1}_{0}\frac{1}{1+nx^2}dx=\frac{\pi}{4}<2.$$

So we get $$\frac{1}{2}<L<2.$$

  • $\begingroup$ the limit L is certainly less than 2 (as DXT has proved above) but who to prove that it is greater than 0.5 $\endgroup$ – Snmohith Raju Apr 16 at 16:44

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