# 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.

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Is the downvote because of self-answering? – Jayesh Badwaik Feb 1 '13 at 18:08
Probably. Why did you post a question to which you already knew the answer? – Todd Wilcox Feb 1 '13 at 18:09
A few reasons 1. It gives me oppurtunity to verify that the solution is indeed correct. (I self study, so even though I get an answer and I am pretty sure about it, there is no real way to verify the solution completely.) 2. It allows probably other people to offer me better solutions. 3. I can do so on a blog, but then it might not get the same attention on the blog. 4. MSE's editing capabilities are better than almost all other blogging software I have found. – Jayesh Badwaik Feb 1 '13 at 18:10
I didn't downvote either, but it seems stylistically and logically wrong: the question is "is the following I reproduce in the question text correct?", the answer "yes/no" with reason. I jabs answered my own questions, when I didn't get answers or, after my thought, believed I could improve on other posting; answering your own questions isn't wrong per se. It's also simply more convenient to have the actual question on top. – gnometorule Feb 1 '13 at 18:42
+1: I see no problem posting a question for which you already know the answer and asking for alternative approaches. – Mike Spivey Feb 1 '13 at 21:02

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

This converges to ${1\over 1+1}={1\over 2}$ as $n\to\infty$.

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Thanks. A different solution. :-) – Jayesh Badwaik Feb 1 '13 at 18:37
+1. Order statistics and expectations... nice! :) – Mike Spivey Feb 1 '13 at 21:03
@MikeSpivey Yeah, it's the way I think. Some of the other answers here are great, too. – Byron Schmuland Feb 1 '13 at 21:06
@Byron: If I may say so, I often enjoy your answers that use probability and statistics to answer questions that appear to be outside of those fields. They are close enough to the way I think that I can appreciate them, but I often wouldn't think of those arguments myself. – Mike Spivey Feb 1 '13 at 21:09
@MikeSpivey Thanks for the kind words. I learn a lot from your answers, too. – Byron Schmuland Feb 1 '13 at 21:12

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}

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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.

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What is the name or proof of this basic result? – Alex Feb 1 '13 at 19:53
@Alex: I don't know if it has a name. Our teacher told us once that the result may be easily proved with the calculus knowledge in high school. This link will help you: math.stackexchange.com/questions/168163/… – OFFSHARING Feb 1 '13 at 20:01
Nice solution Chris'ssister! :-) Your question and the corresponding answers provide a decent tool. :-) – Jayesh Badwaik Feb 1 '13 at 20:32
@OrangeHarvester: Thanks! :-) are you Jayesh? (sister here now) – OFFSHARING Feb 1 '13 at 20:44
Yes. I am Jayesh. Name changed for a month. – Jayesh Badwaik Feb 1 '13 at 22:15

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

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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}

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What do you mean by $I(n) = e^{\log(x)n}$ in the last line? – Antonio Vargas Feb 1 '13 at 18:15
Added the explanation. I hope its correct. – Jayesh Badwaik Feb 1 '13 at 18:23
It is not; you can't just pull a $\log x$ out of the integral since you're integrating with respect to $x$. – Antonio Vargas Feb 1 '13 at 18:29
Arrrrgh. Thanks. – Jayesh Badwaik Feb 1 '13 at 18:30
$I(n)\leq \int_0^1 x^n dx={1\over n+1}.$ – Byron Schmuland Feb 1 '13 at 19:19