Find the limit of the sequence involving integral 
Let $f:[0,1] \rightarrow [0,1]$ increasing function and
$a_n=\int_{0}^{1} \frac {1+(f(x)^n}{1+(f(x))^{n+1}} dx \tag 1$
Prove $a_n$ is convergent and find the limit.


It's easy to prove $a_n \ge 1$ and $a_n$ is decreasing, therefore is convergent. By taking $f$ identical zero, we get $a_n=1, \forall n$ hence the limit is $1$. If the limit is not dependent on $f$ then I have to prove the limit is $1$ for all functions, but I can't figure out how to do it. 
 A: A different approach, not using dominated convergence:
First, I claim that $r^n - r^{n+1} \le \frac{(n-1)^{n-1}}{n^n} $ for all $r\in [0,1]$. To see this, we note that $r^n - r^{n+1} = (n-1)^{n-1}\left(\frac{r}{n-1}\right)^{n-1}(r-1)$, and by AM-GM
$$ \left(\frac{r}{n-1}\right)^{n-1}(1-r) = \left(\sqrt[n]{\underbrace{\left(\frac{r}{n-1}\right)\dots\left(\frac{r}{n-1}\right)}_{n-1}(1-r)}\right)^n\le\left(\frac{\overbrace{\frac{r}{n-1}+\dots+\frac{r}{n-1}}^{n-1} + 1-r}{n}\right)^n = \frac{1}{n^n}.$$
Now
$$\frac{1+f(x)^n}{1+f(x)^{n+1}} = \frac{1+f(x)^{n+1} + f(x)^n - f(x)^{n+1}}{1+f(x)^{n+1}} = 1 + \frac{f(x)^n - f(x)^{n+1}}{1+f(x)^{n+1}}$$
and hence
\begin{align} \int\limits_{0}^{1}{\left(\frac{1+f(x)^n}{1+f(x)^{n+1}}-1\right)\text{ d}x} &= \int\limits_{0}^{1}{\frac{f(x)^n-f(x)^{n+1}}{1+f(x)^{n+1}}\text{ d}x} \\
&\le\int\limits_{0}^{1}{\frac{\frac{(n-1)^{n-1}}{n^n}}{1}\text{ d}x} \\
&=\frac{(n-1)^{n-1}}{n^n}.
\end{align}
Since $\frac{(n-1)^{n-1}}{n^n}\le\frac{n^{n-1}}{n^n} = \frac{1}{n}$ and $\frac{1}{n}\rightarrow 0$ as $n\rightarrow\infty$, it follows that
$$\limsup\limits_{n\rightarrow\infty}{(a_n - 1)}\le 0.$$
Since, as noted in the question, $a_n\ge 1$ for all $n$, it follows that $\lim\limits_{n\rightarrow\infty}{a_n} = 1$, as desired.
A: Hint: Show that the function converges pointwise to 1 and use dominated convergence (or Fatou)
