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?
 A: 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 
$$\sqrt{n}I_n \stackrel{n\to \infty}{\longrightarrow}\frac{\sqrt{\pi}}{2}$$
A: Assuming $n\geq 2$,
$$a_n=\int_{0}^{1}\frac{dx}{(x^2+1)^n} = \int_{0}^{\pi/4}\left(\cos\theta\right)^{2n-2}\,d\theta\leq \int_{0}^{\pi/4}e^{-(n-1)\theta^2}\,d\theta\leq \int_{0}^{+\infty}e^{-(n-1)\theta^2}\,d\theta=\frac{\sqrt{\pi}}{2\sqrt{n-1}} $$
since $\cos(x)\leq e^{-x^2/2}$ can be deduced by integrating $\tan(x)>x$. On the other hand the convexity of $\tan(x)$ implies that over $\left[0,\frac{\pi}{4}\right]$ we have $\log\cos(x)\leq -\frac{x^2}{3}$, so
$$ a_n \geq \int_{0}^{\pi/4}e^{-\frac{3}{2}(n-1)\theta^2}\,d\theta=\frac{\sqrt{\pi}}{\sqrt{6(n-1)}}-\int_{\pi/4}^{+\infty}e^{-\frac{3}{2}(n-1)\theta^2}\,d\theta $$
where
$$\int_{\pi/4}^{+\infty}e^{-\frac{3}{2}(n-1)\theta^2}\,d\theta\leq \int_{\pi/4}^{+\infty}e^{-\frac{3\pi}{8}(n-1)\theta}\,d\theta=o\left(\frac{1}{n}\right).$$
This gives that $\lim_{n\to +\infty}n^{\alpha}a_n$ equals $+\infty$ for any $\alpha>\frac{1}{2}$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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When $\ds{n \to \infty}$, the main contribution to the integral comes from a right neighborhood of $\ds{x = 0}$. Namely, with
Laplace Method
\begin{align}
&\bbox[15px,#ffc]{\lim_{n \to \infty}\bracks{n^{2}\int_{0}^{1}
{\dd x \over \pars{1 +x^{2}}^{n}}}} =
\lim_{n \to \infty}\bracks{n^{2}\int_{0}^{1}
\expo{-n\ln\pars{1 + x^{2}}}\dd x}
\\[5mm] = &\
\lim_{n \to \infty}\pars{n^{2}\int_{0}^{\infty}\expo{-nx^{2}}\dd x} =
\lim_{n \to \infty}\pars{n^{2}\,{\root{\pi} \over 2n^{1/2}}} =\
\bbox[15px,#ffc,border:1px solid navy]{+\infty}
\end{align}
