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.

Solution 1:

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}

Solution 2:

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.

Solution 3:

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: $$s_n \approx \frac{1}{2} + \frac{1}{4 n} - \frac{1}{8 n^3} + ...$$

Solution 4:

Using the substitution $x\mapsto x^{1/n}$ and Dominated Convergence, $$ \begin{align} \lim_{n\to\infty}\int_0^1\frac{nx^{n-1}}{1+x}\,\mathrm{d}x &=\lim_{n\to\infty}\int_0^1\frac1{1+x^{1/n}}\,\mathrm{d}x\\ &=\int_0^1\frac12\,\mathrm{d}x\\ &=\frac12 \end{align} $$