Convergence of the sequence $a_n=\int_0^1{nx^{n-1}\over 1+x}dx$ How to prove the following sequence converges to $0.5$ ?
$$a_n=\int_0^1{nx^{n-1}\over 1+x}dx$$
What I have tried:
I calculated the integral $$a_n=1-n\left(-1\right)^n\left[\ln2-\sum_{i=1}^n {\left(-1\right)^{i+1}\over i}\right]$$
I also noticed ${1\over2}<a_n<1$ $\forall n \in \mathbb{N}$.
Then I wrote a C program and verified that $a_n\to 0.5$ (I didn't know the answer before) by calculating $a_n$ upto $n=9990002$ (starting from $n=2$ and each time increasing $n$ by $10^4$). I can't think of how to prove $\{a_n\}$ is monotone decreasing, which is clear from direct calculation.
 A: Using integration by parts, we obtain
\begin{align}
\int_0^1 \frac{nx^{n-1}}{1+x}dx &=\left.\frac{x^n}{1+x}\right|_0^1+\int_0^1\frac{x^n}
{(1+x)^2}dx
=\frac{1}{2}+r_n,
\end{align}
where clearly
$$
0<r_n\le \int_0^1 x^n\,dx=\frac{1}{n+1}\longrightarrow 0,
$$
as $n\to\infty$.
A: Thinking about the graph of $x^n$ on $[0,1]$ we observe that it stays near $0$ and then sharply jumps to $1$. As such, it makes sense to break up the integral into $[0,c)$ and $[c,1]$ (for some $c$ to be chosen later).
$$ 
a_n = \int_0^c{\frac{n x^{n-1}}{x+1}dx} + \int_c^1{\frac{n x^{n-1}}{x+1}dx} \leq \int_0^c{n x^{n-1}dx} + \int_c^1{\frac{n x^{n-1}}{c+1}dx}\\
= c^n + \frac{1 - c^n}{c+1}.
$$
Now observe that for any fixed $c < 1$, $c^n + \frac{1 - c^n}{c+1} \rightarrow 1/(c+1)$ as $n \rightarrow \infty$. Thus we have $\limsup_{n \rightarrow \infty}{a_n} \leq 1/(c+1)$, and now letting $c \rightarrow 1$ from below we conclude $\limsup_{n \rightarrow \infty}{a_n} \leq 1/2$.
On the other hand $a_n = \int_0^1{\frac{n x^{n-1}}{x+1}dx} \geq \int_0^1{\frac{n x^{n-1}}{2}dx} = 1/2$ for all $n$, so that $\liminf_{n \rightarrow \infty}{a_n} \geq 1/2$.
Thus $$\lim_{n \rightarrow \infty}{a_n} = \frac{1}{2}.$$
A: $\newcommand{\+}{^{\dagger}}%
 \newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
 \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}%
 \newcommand{\dd}{{\rm d}}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\fermi}{\,{\rm f}}%
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
 \newcommand{\half}{{1 \over 2}}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
 \newcommand{\ol}[1]{\overline{#1}}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}%
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
 \newcommand{\sech}{\,{\rm sech}}%
 \newcommand{\sgn}{\,{\rm sgn}}%
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
When $n \gg 1$ the main contribution to the integral comes from $x \sim 1$. Then, we set the change of variables $x = 1 -\epsilon$:
\begin{align}
\color{#0000ff}{\large a_{n}} &= \int_{0}^{1}{nx^{n - 1} \over 1 + x}\,\dd x
=
\half\,n\int_{0}^{1}{\pars{1 - \epsilon}^{n - 1} \over 1 - \epsilon/2}\,\dd\epsilon
\\[3mm]&=
\half\,n\int_{0}^{1}\exp\pars{\bracks{n - 1}\ln\pars{1 - \epsilon} - \ln\pars{1 - {\epsilon \over 2}}}\,\dd\epsilon\quad
{\Large\stackrel{n \gg 1}{\sim}}\quad
\half\,n\int_{0}^{\infty}\exp\pars{-\bracks{n - {3 \over 2}}\epsilon}\,\dd\epsilon
\\[3mm]&=
\half\,{n \over n - 3/2}\quad \color{#0000ff}{\large\stackrel{n \to \infty}{\Huge \to} \quad\half}
\end{align}
A: Define $I_n =\displaystyle \int_0^1 \frac{x^n}{1+x}$. Then you can obtain immediately that $I_{n+1}+I_n = \displaystyle \frac{1}{n+1}$. Next note that $0\leq I_{n+1}\leq I_n$ since for $0\leq x \leq 1$ the inequality $0\leq \frac{x^{n+1}}{1+x} \leq \frac{x^n}{1+x}$ holds.
Therefore $I_n \to 0$ as $n \to \infty$. Thus we have
$$ a_{n+1}+a_n = (n+1)I_{n}+nI_{n-1} = 1+I_n \to 1 $$
Now if you prove that $a_n$ converges, you are done, since $a_{n+1}+a_n \to 1$.
(maybe this is more intricate...)
A: Male the u-substitution $u=x^n$, them apply the dominated convergence theorem. 
A: We have
$$
a_n=\int_0^1\frac{nx^{n-1}}{1+x}\,dx=\frac{x^n}{1+x}\Big|_0^1+\int_0^1\frac{x^n}{(1+x)^2}\,dx=\frac12+\int_0^1\frac{x^n}{(1+x)^2}\,dx \quad \forall n \ge 1.
$$
Since
$$
\int_0^1\frac{x^n}{(1+x)^2}\,dx\le \int_0^1x^n\,dx=\frac{1}{n+1} \quad \forall n\ge 1,
$$
it follows that
$$
\lim_n\int_0^1\frac{x^n}{(1+x)^2}\,dx=0.
$$
Thus $\lim_na_n=\frac12$.
A: EDIT: I feel kind of stupid for not thinking of the easier ways in other posts, but I think this method is kind of cool.
I apologize in advance, this is a lot of math and few words.
$$\begin{align} \int_0^1\frac{nx^{n-1}}{1+x}\text dx&=\int_0^1nx^{n-1}\sum_{k=0}^\infty(-x)^k\text dx\\
&=\sum_{k=0}^\infty \int_0^1nx^{n-1+k}(-1)^k\text dx\\
&=\sum_{k=0}^\infty\frac{(-1)^kn}{n+k} \end{align}$$
Now, you want $$\begin{align}\lim_{n\to\infty}n\sum_{k=0}^\infty \frac{(-1)^k}{n+k}&=\lim_{n\to\infty}n\left(\sum_{k=0}^\infty \frac{1}{n+2k}-\frac{1}{n+2k+1}\right)\\
&=\lim_{n\to\infty}n\sum_{k=0}^\infty \frac1{(n+2k)^2+n+2k}\\
&=\lim_{n\to\infty}n\sum_{k=0}^\infty\frac{1}{(n+2k)^2}\tag 1\\
&=\lim_{n\to\infty}\frac1n\sum_{k=0}^\infty \frac{1}{(1+2\frac kn)^2}\\
&=\int_0^\infty \frac{1}{(1+2x)^2}\text dx\tag 2\\
&=\frac12\int_0^\infty \frac1{(1+x)^2}\text dx\\
&=\frac12\left.\left(-\frac1{x+1}\right)\right|_0^\infty\\
&=\frac12\end{align}$$
$(1)$ is obtained by realizing that $n+2k$ is negligible in comparison to $(n+2k)^2$ as $n$ approaches $\infty$
$(2)$ uses the well-known identity $$\lim_{n\to\infty}\frac1n\sum_{k=an}^{bn}f\left(\frac kn\right)=\int_a^bf(x)\text dx$$
A: There is another way of adressing the problem (but it could be totally out of scope depending of what you are supposed to use for this).  
It can be established that
a(n) = (n / 2) (-PolyGamma[0, n/2] + PolyGamma[0, (1 + n)/2])
Now, a Taylor development of a(n) built around infinity gives an approximation which is
1/2 + 1/(4 n) - 1/(8 n^3)
