Vtgyjfy

Let $s_n$ be a sequence defined as given below for $n geq 1$. Then find out $limlimits_{n to
infty} s_n$.
begin{align}
s_n = intlimits_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 &= intlimits_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+xright)^2} left(int nx^{n-1} d xright) d x
right]^1_0 \
&= left[frac{1}{1+x} int nx^{n-1} d xright]^1_0
- left[int frac{1}{left(1+xright)^2} left(int nx^{n-1} dxright) d xright]^1_0 \
&= left[frac{x^n}{1+x}right]^1_0
- left[int frac{x^n}{left(1+xright)^2} d xright]^1_0 \
&= frac{1}{2} - intlimits_0^1 frac{x^n}{left(1+xright)^2} d x \
end{align}

Now we estimate the remaining integral in the expression
begin{align}
I(n) &= intlimits_0^1 frac{x^n}{left(1+x right)^2} d x \
&leq intlimits_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}
limlimits_{n to infty} s_n = frac{1}{2}
end{align}

We use a basic result in calculus, namely $lim_{nto infty}nint_0^1x^nf(x) dx=f(1)$, $f$ continuous on $[0,1]$
$$lim_{nto infty}left(frac{n}{n-1}times (n-1)intlimits_0^1 x^{n-1} frac{1}{(1+x)} dxright)=frac{1}{2}$$

Chris.

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 $0leq xleq 1$.
Now let $M=max(X_1,dots, X_n)$; its density function is
$$f_M(x)=n F(x)^{n-1}f_X(x)=n,x^{n-1}text{ for }0leq xleq 1.$$
Also, it is not hard to see that $Mto 1$ in distribution as $ntoinfty$.
Now $$int_0^1 {n x^{n-1}over 1+x} ,dx =int_0^1 {1over 1+x}, f_M(x) ,dx
=mathbb{E}left({1over 1+M}right).$$

This converges to
${1over 1+1}={1over 2}$ as $ntoinfty$.

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) $s_n = nint_0^1 frac{x^{n-1}}{1+x} dx < nint_0^1 x^{n-1} dx = 1$

(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) $implies s = lim_{ntoinfty} s_n$ 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:
$$s_n approx frac{1}{2} + frac{1}{4 n} - frac{1}{8 n^3} + ...$$