Consider $$ L = \lim_{n\to \infty} \sqrt n\int_{0}^1 \frac{1}{(1+x^2)^n}dx$$
Given is that $$L < 1/2$$
What is the minimum upper bound on L.
What have i tried :
Generating a recurrence relation for the involved integral : I however could not find a proper line of reasoning to calculate the minimum upper bound.
Finding the value of integral involved for n= 1, 2, 3... and guess a pattern for n. The value for n=1 , 2 , 3 however does not offer an easily observable pattern. . It finally struck me that if i change the denominator of the involved integral from $1+x^2$ to $x^2$ I can argue as follows $$\sqrt n\int_{0}^1 \frac{1}{(1+x^2)^n}dx < \sqrt n \int_{0}^1 \frac{1}{x^{2n}}dx$$ This however did not help me as the integral on the RHS could not be evaluated at the lower limit.I then replaced $(1+ x^2)$ by $(1+x)^2$ As this will also ensure the following: $$\sqrt n \int_{0}^1 \frac{1}{(1+x^2)^n}dx < \sqrt n \int_{0}^1 \frac{1}{(1+x)^{2n}}dx$$ The RHS integral had the value : $$\sqrt n\frac{1}{-2n+1}(\frac{1}{2^{2n-1}}-1)$$ Imposing the limit n goes to infinity on this gives $0$ which is certainly not what I expected. . What am i missing here ?