Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
I find the $(-1)^n$ to be troubling... How did you calculate the integral? (not saying it's wrong, but it just looks funny to me) –  anorton Dec 21 '13 at 20:55
    
I broke up (1+x)^-1 into infinite series then integrated and considered cases - n is even or odd and used ln2=sum of alternating harmonic series. –  Jack's wasted life Dec 21 '13 at 20:58

8 Answers 8

up vote 17 down vote accepted

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 $\lim_na_n=\frac12$.

share|improve this answer

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 $$\lim_{n\to\infty}\frac1n\sum_{k=an}^{bn}f\left(\frac kn\right)=\int_a^bf(x)\text dx$$

share|improve this answer
    
You're right.It is kind of cool.I would've upvoted if I had enough rep...:} –  Jack's wasted life Dec 21 '13 at 21:18

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 $n\to\infty$.

share|improve this answer

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 $$\lim_{n \rightarrow \infty}{a_n} = \frac{1}{2}.$$

share|improve this answer

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ 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}

share|improve this answer

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...)

share|improve this answer

Male the u-substitution $u=x^n$, them apply the dominated convergence theorem.

share|improve this answer

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)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.