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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.

  • Is the downvote because of self-answering? –  Feb 01 '13 at 18:08
  • Probably. Why did you post a question to which you already knew the answer? – Todd Wilcox Feb 01 '13 at 18:09
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    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.
     –  Feb 01 '13 at 18:10
  • That's not always seen as a good thing for Math.SE. See: http://meta.math.stackexchange.com/questions/4680/is-answering-own-question-okay – Todd Wilcox Feb 01 '13 at 18:15
  • @ToddWilcox Counterpoint. http://meta.math.stackexchange.com/q/2244/14082 –  Feb 01 '13 at 18:28
  • I did not downvote, I was offering a possible reason why someone might have. Notice I wrote "not always seen...", by which I meant some people won't like you asking a question to which you already know the answer and therefore probably one of those people downvoted your question. I'm not advocating any position on the subject one way or the other. – Todd Wilcox Feb 01 '13 at 18:31
  • Okay. No problems. :-) Generally, when people downvote, and the OP asks the reason, it is expected that the person who downvoted leave a comment about it. And hence, my assumption. –  Feb 01 '13 at 18:33
  • Sadly I think there are many who downvote without comment, regardless of whether an explanation is requested or not. – Todd Wilcox Feb 01 '13 at 18:34
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    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 01 '13 at 18:42
  • It could also be seen as "reputation boosting": answer a question you can answer well, and get upvoted for the answer you knew in advance. – gnometorule Feb 01 '13 at 18:43
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    +1: I see no problem posting a question for which you already know the answer and asking for alternative approaches. – Mike Spivey Feb 01 '13 at 21:02
  • If you are looking for alternate approaches, you might post your approach so that we know how not to answer. – robjohn Jan 10 '15 at 12:33
  • See also: How to integrate this type of integral: $\lim_{n \rightarrow \infty}{\int_ 0^1}\frac{n x^{n-1}}{1+x}dx$ – Martin Sleziak May 06 '17 at 05:46

5 Answers5

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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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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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user 1591719
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  • What is the name or proof of this basic result? – Alex Feb 01 '13 at 19:53
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    @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: http://math.stackexchange.com/questions/168163/find-functions-family-satisfying-lim-n-to-infty-n-int-01-xn-fx-f1 – user 1591719 Feb 01 '13 at 20:01
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    Nice solution Chris'ssister! :-) Your question and the corresponding answers provide a decent tool. :-) –  Feb 01 '13 at 20:32
  • @OrangeHarvester: Thanks! :-) are you Jayesh? (sister here now) – user 1591719 Feb 01 '13 at 20:44
  • Yes. I am Jayesh. Name changed for a month. –  Feb 01 '13 at 22:15
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Using the substitution $x\mapsto x^{1/n}$ and Dominated Convergence, $$ \begin{align} \lim_{n\to\infty}\int_0^1\frac{nx^{n-1}}{1+x}\,\mathrm{d}x &=\lim_{n\to\infty}\int_0^1\frac1{1+x^{1/n}}\,\mathrm{d}x\\ &=\int_0^1\frac12\,\mathrm{d}x\\ &=\frac12 \end{align} $$

robjohn
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  • Nice Robjohn . I have a doubt . When $\displaystyle \lim_{n\rightarrow \infty} $ and $0<x<1.$. Then $(x)^n=0.$ So $\displaystyle \lim_{n\rightarrow \infty}\int^{1}_{0}\frac{n x^{n-1}}{1+x}dx=0$. please explain me where i am wrong. Thanks – DXT Oct 02 '18 at 07:16
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    @DurgeshTiwari: You cannot simply exchange limits and integrals. There are rules for doing so. Above, I used Dominated Convergence, which says that if $|f_n(x)|\le g(x)$ for all $n$ and $x\in[0,1]$ and $\int_0^1g(x),\mathrm{d}x\lt\infty$, then $$\lim\limits_{n\to\infty}\int_0^1f_n(x),\mathrm{d}x=\int_0^1\lim\limits_{n\to\infty}f_n(x),\mathrm{d}x$$ There is no such $g(x)$ that dominates $\frac{nx^{n-1}}{1+x}$ for all $n$. – robjohn Oct 02 '18 at 08:34
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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: $$s_n \approx \frac{1}{2} + \frac{1}{4 n} - \frac{1}{8 n^3} + ...$$

achille hui
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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. :-) –  Feb 01 '13 at 18:37
  • @MikeSpivey Yeah, it's the way I think. Some of the other answers here are great, too. –  Feb 01 '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 01 '13 at 21:09
  • @MikeSpivey Thanks for the kind words. I learn a lot from your answers, too. –  Feb 01 '13 at 21:12