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Evaluate the following limit : $$\lim_{n\to \infty}\left[(n+1)\int_0^1x^n\ln(1+x)\,dx\right].$$

We have , $$\lim_{n\to \infty}\left[(n+1)\int_0^1x^n\ln(1+x)\,dx\right]$$ $$=\lim_{n\to \infty}\int_0^1\ln(1+x)\,d(x^{n+1})$$Now put , $x^{n+1}=y$. Then , $$=\lim_{n\to \infty}\int_0^1\ln\left(1+y^{\frac{1}{n+1}}\right)\,dy$$Let , $\displaystyle g_n(y)=\ln\left(1+y^{\frac{1}{n+1}}\right)$.

Edit :

Then , $\displaystyle g(y)=\lim_ng_n(y)=\ln 2$ in $(0,1]$.

Now , $\displaystyle \sup_{x\in (0,1]}|g_n(y)-g(y)|=\sup_{x\in (0,1]}\ln\left(\frac{1+y^{\frac{1}{n+1}}}{2}\right)=0$. ( As , $y^{\frac{1}{n+1}}$ is monotone increasing function in $(0,1]$ , so $\ln\left(\frac{1+y^{\frac{1}{n+1}}}{2}\right)$ is also monotone increasing in $(0,1]$ and so it attains its maximum value at $y=1$ . ) So , $\{g_n(y)\}$ converges uniformly to $\ln 2$.

Then ,we can show that $g_n(y)$ converges uniformly to $\ln 2$ in $(0,1]$ and hence the given limit is $\ln 2$.

Is this correct ? Does there any other technique to evaluate the limit ?

Empty
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    "Then ,we can show that $g_n(y)$ converges uniformly to $\ln 2$" Can we? – Did Oct 27 '15 at 20:25
  • @ Did ) I can do it...So I don't think that it is necessary to enhance the length of the post including that analysis. – Empty Oct 27 '15 at 20:29
  • You can do it? I see... But wait, it seems that $g_n(0)=0$ after all. This is strange, don't you think? – Did Oct 27 '15 at 20:31
  • "Then ,we can show that $g_n(y)$ converges uniformly to $\ln 2$ in $(0,1]$" Hmmm... how to say these things? Well, you know, the modified version, "cleverly" replacing $[0,1]$ by $(0,1]$, is as deadly wrong as the previous one. There is no uniform convergence here, so you better start right away finding some other argument. – Did Oct 27 '15 at 20:34
  • @Did) See my Edit in question and tell me what's my mistake to show uniform convergence .... – Empty Oct 28 '15 at 03:34
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    Plotting the graphs of some functions $g_n$ could help... $$\sup_{x\in (0,1]}|g_n(y)-g(y)|=\sup_{x\in (0,1]}\left|\ln\left(\frac{1+y^{\frac{1}{n+1}}}{2}\right)\right|=\left|\ln\left(\frac{1+0^{\frac{1}{n+1}}}{2}\right)\right|=\ln2\ne0$$ – Did Oct 28 '15 at 07:30
  • By the way, I copied the beginning of the formula in my comment from your post and there is a typo: the second supremum should read over $y\in(0,1]$, not over $x\in(0,1]$. – Did Oct 28 '15 at 07:48

3 Answers3

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It seems that you don't actually have uniform convergence, as you claim. In particular, $x^{1/n}$ does not converge to $1$ uniformly on $(0,1]$. However, you have monotone convergence, which is sufficient by the monotone convergence theorem.

Another way to evaluate the limit: with integration by parts, we have $$ \int_0^1(n+1)x^n \ln(x+1)\,dx = (1)^{n+1}\ln(2) - \int_0^1 \frac{x^{n+1}}{x+1}dx $$ And, we have $$ 0 \leq \int_0^1 \frac{x^{n+1}}{x+1}dx \leq \int_0^1 \frac{x^{n+1}}{1}dx = \frac{1}{n+2} $$

Ben Grossmann
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  • "Your method looks fine." Hmmm... See comments on main. – Did Oct 27 '15 at 20:31
  • @Did looks like you're right. Fixed. – Ben Grossmann Oct 27 '15 at 20:36
  • @@Omnomnomnom) See my Edit in question and tell me what's my mistake to show uniform convergence .... I deal with the function $\displaystyle \ln\left(1+y^{\frac{1}{n+1}}\right)$ about uniform convergence , NOT only the function $x^{1/n}$ which is NOT uniform convergent , as you say in your answer. – Empty Oct 28 '15 at 03:37
  • @@ Omnomnomnom ) Please response...where my mistake ? – Empty Oct 28 '15 at 06:55
  • Note that $\ln(1+x)\approx x$ for $x\in (-1,1)$. In any case, Did has the right idea. – Ben Grossmann Oct 28 '15 at 10:59
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Let $U_1,U_2,\dots$ be i.i.d. uniform (0,1) random variables and set $M_n=\max(U_1,\dots, U_{n})$. Then $M_{n+1}$ has density $(n+1) x^n$ for $x\in(0,1)$. As $n\to\infty$ we have $M_{n+1}\to 1$ in distribution, so $$\int_0^1 (n+1) x^n \ln(1+x)\,dx=\mathbb{E}(\ln(1+M_{n+1}))\to \ln(1+1)=\ln(2).$$

Similar solutions using order statistics:

Limit of $s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx$ as $n \to \infty$

Computing $\lim\limits_{n\to+\infty}n\int_{0}^{\pi/2}xf(x)\cos ^n xdx$

Solving $\lim_{n\to\infty}(n\int_0^{\pi/4}(\tan x)^ndx)$?

Community
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\lim_{n\to \infty} \bracks{\pars{n + 1}\int_{0}^{1}x^{n}\ln\pars{1 + x}\,\dd x}} = \lim_{n\to \infty} \bracks{\pars{n + 1}\int_{0}^{1}\pars{1 - x}^{n}\ln\pars{2 - x}\,\dd x} \\[5mm] = &\ \lim_{n\to \infty} \bracks{\pars{n + 1}\int_{0}^{1}\exp\pars{n\ln\pars{1 - x}} \ln\pars{2 - x}\,\dd x} \\[5mm] = &\ \lim_{n\to \infty} \bracks{\pars{n + 1}\int_{0}^{\infty}\exp\pars{-nx}\ln\pars{2}\,\dd x} \qquad\pars{~Laplace'\!s\ Method~} \\[5mm] = &\ \ln\pars{2}\lim_{n \to \infty}\bracks{\pars{n + 1}{1 \over n}} = \bbx{\ln\pars{2}} \end{align}

See Laplace's Method.

Felix Marin
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