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Here is a limit that can be computed directly by performing the integration and then taking
the limit, but the way is rather ugly. What else can we do? Might we avoid the integration? $$\lim_{n\to\infty} \frac{1}{\sqrt{n}}\int_{ 1/{\sqrt{n}}}^{1}\frac{\ln(1+x)}{x^3}\mathrm{d}x$$

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@GitGud: from my brother and some pals. –  Chris's sis Mar 3 '13 at 17:44
Downvote? Who knows? Maybe stating a problem with no effort shown to solve it. –  GEdgar Mar 3 '13 at 18:01
JFYI, salutations and signatures must be removed from posts. Refer to Meta.SO for more information. Please try to be more polite, this is not your personal website. –  Gigili Mar 3 '13 at 18:11
Chris's sister: your contributions here are very worthwhile. I have enjoyed following your questions and answers for a while now. Please don't rage-quit. –  AndrewG Mar 3 '13 at 18:22
I believe every single user in this site aims to know more about Maths, instead of arguing. Please stay calm and rational. –  Michael Li Mar 3 '13 at 18:29

7 Answers 7

up vote 14 down vote accepted

An alternate way is to notice that expanding the integrand also gives:

$$\frac{1}{x^2}-\frac{1}{x}<\frac{\ln(1+x)}{x^3}< \frac{1}{x^2}\;\,\text{for}\;\,0<x<1$$


$$\frac{1}{\sqrt{n}}\int_{1/\sqrt{n}}^1\frac{1}{x^2}-\frac{1}{x}\,dx< \frac{1}{\sqrt{n}}\int_{1/\sqrt{n}}^1\frac{\ln(1+x)}{x^3}dx< \frac{1}{\sqrt{n}}\int_{1/\sqrt{n}}^1\frac{1}{x^2}dx$$

$$1-\frac{1}{\sqrt{n}}-\frac{\ln\sqrt{n}}{\sqrt{n}}<\frac{1}{\sqrt{n}}\int_{1/\sqrt{n}}^1\frac{\ln(1+x)}{x^3}dx< 1-\frac{1}{\sqrt{n}}$$

$$\frac{1}{\sqrt{n}}\int_{1/\sqrt{n}}^1\frac{\ln(1+x)}{x^3}\to 1$$

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That's a very nice way. –  Chris's sis Mar 3 '13 at 19:53
(+1) nice answer. –  Mhenni Benghorbal Mar 5 '13 at 2:04
@MhenniBenghorbal Thanks! –  L. F. Mar 5 '13 at 12:55

By L`Hopital's rule:

$$ \lim_{n\rightarrow \infty} -\frac{\ln(1 + \sqrt{n}) n^{1.5} \frac{-1}{2 n^{1.5}}}{0.5 n^{-0.5}}$$ $$ = \lim_{n\rightarrow \infty} \ln(1 + \frac{1}{\sqrt{n}}) \sqrt{n}$$ Using the fact that $$\ln(1 + x) = x + O(x^2)$$

We see that the limit is $1$.

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Thanks (+1). I think we should also tell a word about the behaviour of the numerator before applying L`Hopital's rule. –  Chris's sis Mar 3 '13 at 17:58
Perhaps, as $x\rightarrow 0$, that integral strongly resembles that of $\int \frac{1}{x^2}$. –  muzzlator Mar 3 '13 at 18:05
@Chris'ssisterandpals it's only required that the denominator tend to $\infty$ to apply L'Hopital's rule. –  Antonio Vargas Mar 3 '13 at 18:05
Ah, didn't know that, thank you antonio. –  muzzlator Mar 3 '13 at 18:05

Not strictly avoiding the integration, but expanding the integrand makes it pretty straightforward:


$$\int_{\frac{1}{\sqrt{n}}}^1 \frac{\ln (1+x)}{x^3}dx=\left[-\frac{1}{x}-\frac{\ln x}{2}+\frac{x}{3}-\cdots\right]_{1/\sqrt{n}}^1=-\frac{3}{4}+\sqrt{n}+\frac{\ln \sqrt{n}}{2}-\frac{1}{3\sqrt{n}}+\cdots$$

$$\frac{1}{\sqrt{n}}\int_{\frac{1}{\sqrt{n}}}^1 \frac{\ln (1+x)}{x^3}dx=1-\frac{3}{4\sqrt{n}}+\frac{\ln \sqrt{n}}{2\sqrt{n}}-\frac{1}{3n}+\cdots\to 1$$

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That was pretty fast. Thanks! –  Chris's sis Mar 3 '13 at 18:18

$$\lim_{n\to \infty}\dfrac{1}{\sqrt{n}}\int_{\frac{1}{\sqrt{n}}}^1\dfrac{\ln(1+x)}{x^3}\mathrm dx=\lim _{t\to \infty}\frac{1}{t}\int_{\frac{1}{t}}^1\dfrac{\ln(1+x)}{x^3}\mathrm dx= \lim _{t\to \infty}\ln(1+t^{-1})t^3/t^2=1$$ By l'Hospital's rule and $\log(1+x)/x\to1$

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Close to $x = 0$, we have $$\frac{\log(1+x)}{x^3} \sim \frac{1}{x^2} - \frac{1}{2x} + \cdots$$ Pick a $\delta > 0$ such that the error term in R.H.S is $O(1)$ on $(0,\delta)$, we have:

$$\begin{align} \int_{\frac{1}{\sqrt{n}}}^1 \frac{\log(1+x)}{x^3} dx &= \left(\int_{\frac{1}{\sqrt{n}}}^\delta + \int_{\delta}^1\right) \frac{\log(1+x)}{x^3} dx\\ &= \left[ -\frac{1}{x} -\frac12 \log x \right]_{\frac{1}{\sqrt{n}}}^\delta + O(\delta) + \int_{\delta}^1 \frac{\log(1+x)}{x^3} dx\\ &= \sqrt{n} +\frac12 \log ( \sqrt{n} ) + O(\frac{1}{\delta}) \end{align}$$ Notice $\delta$ has be chosen independent of $n$, we have: $$\lim_{n\to\infty} \frac{1}{\sqrt{n}} \int_{\frac{1}{\sqrt{n}}}^1 \frac{\log(1+x)}{x^3} dx = \lim_{n\to\infty} \frac{1}{\sqrt{n}} \left(\sqrt{n} +\frac12 \log ( \sqrt{n} ) + O(\frac{1}{\delta})\right) = 1$$

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Thanks for your way (+1) Always rigorous. –  Chris's sis Mar 3 '13 at 20:20

We have with the fact $x-\frac{x^2}{2}\leq\log(1+x)\leq x$, $$\lim_{n\to \infty}\frac{1}{\sqrt{n}}\int_{\frac{1}{\sqrt{n}}}^1\frac{\log(1+x)}{x^3}\leq\lim_{n\to \infty}\frac{1}{\sqrt{n}}\int_{\frac{1}{\sqrt{n}}}^1\frac{1}{x^2}=\lim_{n\to \infty}\frac{1}{\sqrt{n}}({\sqrt{n}}-1)=1.$$ and $$1=\lim_{n\to \infty}\frac{1}{\sqrt{n}}(({\sqrt{n}}-1)-\frac{1}{2}\log\sqrt{n})=\lim_{n\to \infty}\frac{1}{\sqrt{n}}\int_{\frac{1}{\sqrt{n}}}^1\frac{1}{x^2}-\frac{1}{2x}\leq\lim_{n\to \infty}\frac{1}{\sqrt{n}}\int_{\frac{1}{\sqrt{n}}}^1\frac{\log(1+x)}{x^3}$$ We can conclude.

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There's a mistake in your integration. –  Antonio Vargas Mar 3 '13 at 18:09
@AntonioVargas Thanks. –  Sami Ben Romdhane Mar 3 '13 at 18:38
Thanks for your way (+1) –  Chris's sis Mar 3 '13 at 20:20

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

\begin{align} &\color{#00f}{\large\lim_{n \to \infty}\bracks{{1 \over \root{n}}% \int_{ 1/\root{n}}^{1}{\ln\pars{1 + x} \over x^{3}}\,\dd x}} \\[3mm]&=\lim_{n \to \infty}\bracks{{1 \over \root{n}}% \int_{ 1/\root{n}}^{1}{\ln\pars{1 + x} - x + x^{2}/2 \over x^{3}}\,\dd x +{1 \over \root{n}}% \int_{ 1/\root{n}}^{1}{x - x^{2}/2 \over x^{3}}\,\dd x} \\[3mm]&=\lim_{n \to \infty}\bracks{{1 \over \root{n}}% \int_{ 1/\root{n}}^{1}{\ln\pars{1 + x} - x + x^{2}/2 \over x^{3}}\,\dd x + 1 - {1 \over \root{n}} - {\ln\pars{n} \over 4\root{n}}} \\[3mm]&=1 + \lim_{n \to \infty}\bracks{{1 \over \root{n}}% \int_{ 1/\root{n}}^{1}{\ln\pars{1 + x} - x + x^{2}/2 \over x^{3}}\,\dd x - {1 \over \root{n}} - {\ln\pars{n} \over 4\root{n}}} =\color{#00f}{\Large 1} \end{align}

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