Define $I_n=\int_0^1\frac{x^n}{\sqrt{x^2+1}}dx$ for every $n\in\mathbb{N}$. Prove that $\lim_{n\to\infty}nI_n=\frac{1}{\sqrt 2}$. 
Question: Define $I_n=\int_0^1\frac{x^n}{\sqrt{x^2+1}}dx$ for every $n\in\mathbb{N}$. Prove that $$\lim_{n\to\infty}nI_n=\frac{1}{\sqrt 2}$$. 

My approach: Given that $I_n=\int_0^1\frac{x^n}{\sqrt{x^2+1}}dx, \forall n\in\mathbb{N}.$ Let us make the substitution $x^n=t$, then $$nI_n=\int_0^1\frac{dt}{\sqrt{1+t^{-2/n}}}.$$
Now since $0\le t\le 1\implies \frac{1}{t}\ge 1\implies \left(\frac{1}{t}\right)^{2/n}\ge 1 \implies 1+\left(\frac{1}{t}\right)^{2/n}\ge 2\implies \sqrt{1+\left(\frac{1}{t}\right)^{2/n}}\ge \sqrt 2.$ 
This implies that $$\frac{1}{\sqrt{1+\left(\frac{1}{t}\right)^{2/n}}}\le\frac{1}{\sqrt 2}\\ \implies \int_0^1 \frac{dt}{\sqrt{1+\left(\frac{1}{t}\right)^{2/n}}}\le \int_0^1\frac{dt}{\sqrt 2}=\frac{1}{\sqrt 2}.$$
So, as you can see, I am trying to solve the question using Sandwich theorem. 
Can someone help me to proceed after this? 
Also, in $$\lim_{n\to\infty}nI_n=\lim_{n\to\infty}\int_0^1\frac{dt}{\sqrt{1+t^{-2/n}}},$$ the limit and integral interchangeable?
 A: Actually you are done. You already have: $$nI_n=\int_0^1\frac{dt}{\sqrt{1+t^{-2/n}}}$$ hence $$\begin{align}\lim_{n\to\infty} nI_n &= \lim_{n\to\infty}\int_0^1\frac{dt}{\sqrt{1+t^{-2/n}}} \\ &= \int_0^1 \lim_{n\to\infty} \frac{dt}{\sqrt{1+t^{-2/n}}} \\ &=  \int_0^1\frac{dt}{\sqrt{1+1}} = \frac{1}{\sqrt{2}} \end{align}$$
You can exchange limit and integration due to dominated convergence theorem.
A: You can still use Sandwich theorem, if you haven't yet seen dominated convergence theorem :
Let $ n $ be a positive integer.
As you said, using the substitution $ \left\lbrace\begin{aligned}y&=x^{n}\\ \mathrm{d}y &=n x^{n-1}\,\mathrm{d}x\end{aligned}\right. $, we get : $$\int_{0}^{1}{\frac{n x^{n}}{\sqrt{1+x^{2}}}\,\mathrm{d}x}=\int_{0}^{1}{\frac{y^{\frac{1}{n}}}{\sqrt{1+y^{\frac{2}{n}}}}\,\mathrm{d}y}$$
Meaning, we have : \begin{aligned}\left|\frac{1}{\sqrt{2}}-n I_{n}\right|&=\left|\int_{0}^{1}{\left(\frac{1}{\sqrt{2}}-\frac{y^{\frac{1}{n}}}{\sqrt{1+y^{\frac{2}{n}}}}\right)\mathrm{d}y}\right| \\ &=\left|\int_{0}^{1}{\frac{\sqrt{1+y^{\frac{2}{n}}}-\sqrt{2}y^{\frac{1}{n}}}{\sqrt{2\left(1+y^{\frac{2}{n}}\right)}}\,\mathrm{d}y}\right|\\ &=\int_{0}^{1}{\frac{1-y^{\frac{2}{n}}}{\sqrt{2\left(1+y^{\frac{2}{n}}\right)}\left(\sqrt{1+y^{\frac{2}{n}}}+\sqrt{2}y^{\frac{1}{n}}\right)}\,\mathrm{d}y}\end{aligned}
Since $ \left(\forall y\in\left[0,1\right]\right),\ \sqrt{2\left(1+y^{\frac{2}{n}}\right)}\left(\sqrt{1+y^{\frac{2}{n}}}+\sqrt{2}y^{\frac{1}{n}}\right)\geq \sqrt{2}+\sqrt{2}y^{\frac{2}{n}}\geq 1 $, we have : $ \int\limits_{0}^{1}{\frac{1-y^{\frac{2}{n}}}{\sqrt{2\left(1+y^{\frac{2}{n}}\right)}\left(\sqrt{1+y^{\frac{2}{n}}}+\sqrt{2}y^{\frac{1}{n}}\right)}\,\mathrm{d}y}\leq\int\limits_{0}^{1}{\left(1-y^{\frac{2}{n}}\right)\mathrm{d}y} $, and thus : $$ \left|\frac{1}{\sqrt{2}}-n I_{n}\right|\leq\int_{0}^{1}{\left(1-y^{\frac{2}{n}}\right)\mathrm{d}y}=\frac{2}{n+2}\underset{n\to +\infty}{\longrightarrow}0 $$
Hence : $$ \lim_{n\to +\infty}{nI_{n}}=\frac{1}{\sqrt{2}} $$
A: you can use the Binomial series
$${{\left( 1+{{x}^{2}} \right)}^{-1/2}}=\sum\nolimits_{k=0}^{\infty }{\left( \begin{align}
  & -1/2 \\ 
 & \ \ k \\ 
\end{align} \right){{x}^{2k}}}$$
$${n{I}_{n}}=n\int_{0}^{1}{{{x}^{n}}{{\left( 1+{{x}^{2}} \right)}^{-1/2}}dx}=n\sum\nolimits_{k=0}^{\infty }{\left\{ \left( \begin{align}
  & -1/2 \\ 
 & \ \ k \\ 
\end{align} \right)\int_{0}^{1}{{{x}^{2k+n}}dx} \right\}}=\sum\nolimits_{k=0}^{\infty }{\frac{n\left( \begin{align}
  & -1/2 \\ 
 & \ \ k \\ 
\end{align} \right)}{\left( 2k+n+1 \right)}}$$
$$\underset{n\to \infty }{\mathop{\lim }}\,{n{I}_{n}}=\underset{n\to \infty }{\mathop{\lim }}\,\sum\nolimits_{k=0}^{\infty }{\frac{n\left( \begin{align}
  & -1/2 \\ 
 & \ \ k \\ 
\end{align} \right)}{\left( 2k+n+1 \right)}}=\sum\nolimits_{k=0}^{\infty }{\underset{n\to \infty }{\mathop{\lim }}\,\frac{n\left( \begin{align}
  & -1/2 \\ 
 & \ \ k \\ 
\end{align} \right)}{\left( 2k+n+1 \right)}}=\sum\nolimits_{k=0}^{\infty }{\left( \begin{align}
  & -1/2 \\ 
 & \ \ k \\ 
\end{align} \right)}={{\left( 1+1 \right)}^{-1/2}}=\frac{1}{\sqrt{2}}$$
A: Since $n/(n+1)\to 1$ the desired limit is equal to the limit of $$(n+1)\int_{0}^{1}\frac{x^n}{\sqrt{1+x^2}}\,dx=\left.\frac{x^{n+1}}{\sqrt{1+x^2}}\right|_{x=0}^{x=1}+\int_{0}^{1}\frac{x^{n+2}}{(1+x^2)^{3/2}}\,dx$$ and this is same as $$\frac{1}{\sqrt{2}}+\int_{0}^{1}\frac{x^{n+2}}{(1+x^2)^{3/2}}\,dx$$ The integral above clearly lies between $$\frac{1}{2\sqrt{2}}\int_{0}^{1}x^{n+2}\,dx=\frac{1}{2\sqrt{2}(n+3)}$$ and $$\int_{0}^{1}x^{n+2}\,dx=\frac{1}{n+3}$$ and thus by Squeeze Theorem tends to $0$. Therefore the desired limit is $1/\sqrt{2}$.
The same technique can be used to prove more generally that $$n\int_{0}^{1}x^nf(x)\,dx\to f(1)$$
A: Yet another approach is to use integration by parts to obtain
\begin{equation*}
I_{n} + I_{n-2} = \int_{0}^{1}x^{n-2}\sqrt{1+x^{2}}\,\mathrm{d}x = \frac{\sqrt{2}}{n-1} - \frac{1}{n-1}I_{n}
\end{equation*}
for $n \geq 2$, from which we get
\begin{equation*}
nI_{n} = \sqrt{2} - (n-1)I_{n-2} = \sqrt{2} - (n-2)I_{n-2} - I_{n-2}. \tag*{(1)}
\end{equation*}
Define $S_{n} = nI_{n}$ for each $n\in \mathbb{N}$. You have already shown that the sequence $\{S_{n}\}_{n\geq 2}$ is bounded above by $1/\sqrt{2}$, and note also that
\begin{align}
S_{n+1} - S_{n} &= \int_{0}^{1}\frac{(n+1)x^{n+1}-nx^{n}}{\sqrt{1+x^{2}}}\,\mathrm{d}x \\
&\geq \frac{1}{\sqrt{2}}\int_{0}^{1}\big((n+1)x^{n+1} - nx^{n}\big)\,\mathrm{d}x \\
&= \frac{1}{\sqrt{2}(n+1)(n+2)} > 0
\end{align}
so that the sequence $\{S_{n}\}_{n\geq 2}$ is increasing. Thus by the monotone convergence theorem, $S_{n}$ converges to a limit $L$. It is therefore valid to take the limit as $n \to \infty$ on both sides of $(1)$, noting that $I_{n-2} \to 0$, to find that
\begin{equation*}
L=\sqrt{2} - L
\end{equation*}
or
\begin{equation*}
L = \frac{1}{\sqrt{2}}.
\end{equation*}
