$I_n=\int_0^1{\frac{x^n}{x^n+1}}$. Prove $I_{n+1} \le I_n$ for any $n \in \mathbb N$ $$I_n=\int_0^1{\frac{x^n}{x^n+1}}$$
Prove $\lim_{n\to\infty}{I_n} = 0$
Here is what I tried.
First, I rewrite $I_n$.
$$I_n=\int_0^1{1-\frac{1}{x^n+1}}=1 - \int_0^1{\frac{1}{x^n+1}}$$ 
Now the limit becomes:
$$L=1-\lim_{n\to\infty}\int_0^1{\frac{1}{x^n+1}}$$
Next I try to solve the limit of the integral using Squeeze Theorem, with no success.
Using the fact $0\le x \le 1$ I get to the following double inequality:
$$\int_0^1\frac{1}{x^{n-1}+1} \le \int_0^1\frac{1}{x^{n}+1} \le 1 $$
$$I_{n-1} \le I_n \le 1$$ 
I don't know what to do next, I would greatly appreciate some help.
 A: It seems that your question has been answered in the comments. However you may complete your argument by using the squeezing theorem to the integrals of the following inequalities (similar to Sangchul Lee's comment): 
 $$1-x^n\leq \frac 1{1+x^n}\leq 1,~~0\leq x\leq 1.$$ 
A: For any $\epsilon \gt 0$, $\int_0^1\frac{1}{x^n+1}dx\gt K_n(\epsilon)=\int_0^{1-\epsilon}\frac{1}{x^n+1}dx$. 
For every $\delta \gt 0$, there exists an $N$, so that for all $n\gt N$, $x^n\lt \delta$.
In this case, $K_n(\epsilon)\gt\frac{1-\epsilon}{1+\delta}$.  Since  $\delta$ is arbitrarily small, $\lim_{n\to \infty}K_n(\epsilon)\ge 1-\epsilon$ so $\lim_{n\to \infty}\int_0^1\frac{1}{x^n+1}dx \ge 1-\epsilon$.  Because $\epsilon $ is arbitrary,  $\lim_{n\to\infty}\int_0^1\frac{1}{x^n+1}dx=1$ and $I_n\to 0$.
A: $I_n
=\int_0^1{\frac{x^n}{x^n+1}}dx
=\int_0^1{\frac{x^n+1-1}{x^n+1}}dx
=1-\int_0^1{\frac{1}{x^n+1}}dx
=1-J_n$.
Need to show that
$J_{n+1} \ge J_n
$.
$\begin{array}\\
J_{n+1} - J_n
&=\int_0^1{\frac{1}{x^{n+1}+1}}dx-\int_0^1{\frac{1}{x^n+1}}dx\\
&=\int_0^1(\frac{1}{x^{n+1}+1}-\frac{1}{x^n+1})dx\\
&=\int_0^1\frac{x^n+1-(x^{n+1}+1)}{(x^{n+1}+1)(x^n+1)}dx\\
&=\int_0^1\frac{x^n-x^{n+1}}{(x^{n+1}+1)(x^n+1)}dx\\
&=\int_0^1\frac{x^n(1-x)}{(x^{n+1}+1)(x^n+1)}dx\\
&\gt 0\\
\end{array}
$
More directly:
$\begin{array}\\
I_{n} - I_{n+1}
&=\int_0^1{\frac{x^n}{x^n+1}}dx-\int_0^1{\frac{x^{n+1}}{x^{n+1}+1}}dx\\
&=\int_0^1(\frac{x^n}{x^n+1}-\frac{x^{n+1}}{x^{n+1}+1})dx\\
&=\int_0^1\frac{x^n(x^{n+1}+1)-x^{n+1}(x^n+1)}{(x^n+1)(x^{n+1}+1)}dx\\
&=\int_0^1\frac{x^{2n+1}+x^n-x^{2n+1}-x^{n+1}}{(x^n+1)(x^{n+1}+1)}dx\\
&=\int_0^1\frac{x^n-x^{n+1}}{(x^n+1)(x^{n+1}+1)}dx\\
&=\int_0^1\frac{x^n(1-x)}{(x^n+1)(x^{n+1}+1)}dx\\
&\gt 0\\
\end{array}
$
A: Use the fact that if $f(x)\leq g(x)$ for all $x\in [a, b]$, then we have $\int _{a}^{b} f(x)dx\leq \int_{a}^{b} g(x)dx$,  in this case see that for $x\in [0, 1]$,  we have,  $\frac{x^{n+1}}{x^{n+1}+1}\leq \frac{x^{n}}{x^{n}+1}$,  now just integrate both side. 
