Limit of this expression when n tends to infinity The limit is equal to: $$\lim_{n\to\infty} n^2 \int_0^1 \frac{1}{(1+x^2)^n} dx$$
P.S. What should be the better approach for such kinds of problems?

This might be a bit overkill but at least it works to show the divergence of the sequence and gives a nice additional result:


*

*Consider first $I_n = \int_0^1\frac{dx}{(1+x^2)^n}$.

*To bring in somehow the exponential function susbtitute $nx^2 = y$:
$$I_n =\frac{1}{2\sqrt{n}}\underbrace{\int_0^n\frac{dy}{\sqrt{y}\left(1+\frac{y}{n}\right)^{n}}}_{J_n := }$$

*Now, using DCT (dominated convergence theorem) gives
$$J_n \stackrel{DCT:\;n \to \infty}{\longrightarrow}\int_0^{\infty}\frac{e^{-y}}{\sqrt{y}}dy = \Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}$$
Hence, we see 
$$n^2I_n = n^2\frac{J_n}{2\sqrt{n}}= \frac{n^{\frac{3}{2}}}{2}\underbrace{J_n}_{\stackrel{n\to \infty}{\longrightarrow}\sqrt{\pi}} \stackrel{n\to \infty}{\longrightarrow}\infty$$
p.s.:
The additional nice result is that 
$$\sqrt{n}I_n \stackrel{n\to \infty}{\longrightarrow}\frac{\sqrt{\pi}}{2}$$
