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. 

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}
