Limit of $s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx$ as $n \to \infty$ Let $s_n$ be a sequence defined as given below for $n \geq 1$. Then find out $\lim\limits_{n \to
\infty} s_n$. 
 \begin{align}
  s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx
 \end{align}
I have written a solution of my own, but I would like to know it is completely correct, and also I would like if people post more alternative solutions. 
 A: Notice
 (1) $\frac{s_n}{n} + \frac{s_{n+1}}{n+1} = \int_0^1 x^{n-1} dx = \frac{1}{n} \implies s_n + s_{n+1} = 1 + \frac{s_{n+1}}{n+1}$.
 (2) $s_n = n\int_0^1 \frac{x^{n-1}}{1+x} dx < n\int_0^1 x^{n-1} dx = 1$
 (3) $s_{n+1} - s_n = \int_0^1 \frac{d (x^{n+1}-x^n)}{1+x} = \int_0^1 x^n \frac{1-x}{(1+x)^2} dx > 0$
(2+3) $\implies s = \lim_{n\to\infty} s_n$ exists and (1+2) $\implies s+s = 1 + 0 \implies s = \frac{1}{2}$.
In any event, $s_n$ can be evaluated exactly to $n (\psi(n) - \psi(\frac{n}{2}) - \ln{2})$ where $\psi(x)$ is the diagamma function. Since $\psi(x) \approx \ln(x) - \frac{1}{2x} - \frac{1}{12x^2} + \frac{1}{120x^4} + ... $ as $x \to \infty$, we know:
$$s_n \approx \frac{1}{2} + \frac{1}{4 n} - \frac{1}{8 n^3} + ...$$
A: Using the substitution $x\mapsto x^{1/n}$ and Dominated Convergence,
$$
\begin{align}
\lim_{n\to\infty}\int_0^1\frac{nx^{n-1}}{1+x}\,\mathrm{d}x
&=\lim_{n\to\infty}\int_0^1\frac1{1+x^{1/n}}\,\mathrm{d}x\\
&=\int_0^1\frac12\,\mathrm{d}x\\
&=\frac12
\end{align}
$$
A: Here's a solution based on order statistics, similar to my answer here.
Let $X_1,\dots, X_n$ be i.i.d. uniform(0,1) random variables. 
The distribution function of $X$ is $F(x)=x$ and density $f(x)=1$ for $0\leq x\leq 1$.
Now let $M=\max(X_1,\dots, X_n)$; its density function is 
$$f_M(x)=n F(x)^{n-1}f_X(x)=n\,x^{n-1}\text{ for }0\leq x\leq 1.$$ 
Also, it is not hard to see that  $M\to 1$ in distribution as $n\to\infty$.
Now $$\int_0^1 {n x^{n-1}\over 1+x} \,dx =\int_0^1 {1\over 1+x}\, f_M(x) \,dx
=\mathbb{E}\left({1\over 1+M}\right).$$ 
This converges to 
${1\over 1+1}={1\over 2}$ as $n\to\infty$.
A: We simplify the formulate for $s_n$ by integrating by parts.
\begin{align}
   s_n &= \int\limits_0^1 \frac{nx^{n-1}}{1+x} d x \\
       &= \left[
          \frac{1}{1+x} \int nx^{n-1} d x  
          - \int \frac{1}{\left(1+x\right)^2} \left(\int nx^{n-1} d x\right) d x
          \right]^1_0 \\
       &= \left[\frac{1}{1+x} \int nx^{n-1} d x\right]^1_0 
          - \left[\int \frac{1}{\left(1+x\right)^2} \left(\int nx^{n-1} dx\right) d x\right]^1_0 \\
       &= \left[\frac{x^n}{1+x}\right]^1_0  
          - \left[\int \frac{x^n}{\left(1+x\right)^2} d x\right]^1_0 \\
       &= \frac{1}{2} - \int\limits_0^1 \frac{x^n}{\left(1+x\right)^2} d x \\ 
  \end{align}
Now we estimate the remaining integral in the expression 
  \begin{align}
   I(n) &= \int\limits_0^1 \frac{x^n}{\left(1+x \right)^2} d x \\
        &\leq \int\limits_0^1 x^n d x \\
        &= \frac{1}{n+1}
  \end{align}
Hence, $I(n) \to 0$ as $n \to \infty$. 
And so, the expression can be rewritten as 
  \begin{align}
   \lim\limits_{n \to \infty} s_n = \frac{1}{2}
  \end{align}
A: We use a basic result in calculus, namely $\lim_{n\to \infty}n\int_0^1x^nf(x)  \ dx=f(1)$, $f$ continuous on $[0,1]$
$$\lim_{n\to \infty}\left(\frac{n}{n-1}\times (n-1)\int\limits_0^1 x^{n-1} \frac{1}{(1+x)} dx\right)=\frac{1}{2}$$
Chris.
