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Determine $$\lim_{n\to \infty} n\left(1+(n+1)\ln \frac{n}{n+1}\right)$$ I noticed the indeterminate case $\infty \cdot 0$ and I tried to get them all under the $\ln$, but it got more complicated and I reached another indeterminate form. The same happened when I tried to use Stolz-Cesaro.

EDIT: is there an elementary solution, without l'Hospital or Taylor series?

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    $\begingroup$ Power series estimate $\log(1-x) \simeq -x-\frac{1}{2}x^{2}$ suggests: $$1+(n+1)\log \left(1-\frac{1}{n+1}\right) \simeq 1-(n+1) \left(\frac{1}{n+1}+\frac{1}{2(n+1)^{2}}\right)$$ yielding a limit of $-1/2$ @SBA: thanks, got it $\endgroup$ – preferred_anon Oct 8 '17 at 20:33
  • $\begingroup$ Writing $\ln((n/(n+1))=-\ln(1+1/n)$ makes things simpler. $\endgroup$ – marty cohen Oct 9 '17 at 14:19
  • $\begingroup$ Just wondering, have you understood my answer, since it's the one that you asked for in your edited question? $\endgroup$ – user21820 Oct 10 '17 at 15:09
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Let $\frac {n}{n+1}=e^x.$ As $n\to \infty,\;\; x\to 0$ through negative values.The expression is $$e^x\cdot \frac {1+x-e^x}{(1-e^x)^2}.$$ Applying l'Hopital's Rule to $\frac {1+x-e^x}{(1-e^x)^2}$ we get a limit of $\frac {-1}{2}$ as $x\to 0.$ And the far left term $e^x$ in the expression goes to $1$ as $x\to 0.$

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  • $\begingroup$ The comment by Daniel Littlewood also gives a a quick solution, provided that you are familiar with the theory of power-series expansions. $\endgroup$ – DanielWainfleet Oct 9 '17 at 2:00
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Compute the limit at infinity of $$ f(x)=(x-1)\left(1+x\ln\frac{x-1}{x}\right) $$ which, with $t=1/x$, is the same as $$ \lim_{t\to0^+}\frac{1-t}{t}\left(1+\frac{\ln(1-t)}{t}\right)= \lim_{t\to0^+}\frac{t+\ln(1-t)}{t^2}= \lim_{t\to0^+}\frac{t-t-t^2/2+o(t^2)}{t^2}=-\frac{1}{2} $$ The factor $1-t$ can be disregarded, as it has limit $1$ and the remaining factor has a limit.

The original limit is $$ \lim_{n\to\infty}f(n+1) $$

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Upgrading my comment to an answer since I think I actually believe in it:
A Taylor Series expansion near $x=0$ gives $$\log(1-x)=-x-\frac{1}{2}x^{2}+o(x^{2})$$ as $x \to 0$. Hence, taking $x=\frac{1}{n+1}$, $$1+(n+1)\log\left(1-\frac{1}{n+1}\right)=1-\frac{n+1}{n+1}-\frac{n+1}{2(n+1)^{2}}+o(n^{-1})$$ and, multiplying by $n$, $$n\left(1+(n+1)\log\left(1-\frac{1}{n+1}\right)\right)=-\frac{n}{2(n+1)}+o(1)$$ which safely tends to $-1/2$.

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  • $\begingroup$ This is indeed correct, and I always evaluate limits this way. However, you may also be interested in my answer that does not require knowledge of the full Taylor expansion. =) $\endgroup$ – user21820 Oct 9 '17 at 3:51
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$$\ln\dfrac{n}{n+1}=\ln(1-\dfrac{1}{n+1})\sim-\dfrac{1}{n+1}-\dfrac{1}{2(n+1)^2}-\mathcal{O}(\dfrac{1}{n^3})$$ then $$1+(n+1)\ln\dfrac{n}{n+1}\sim-\dfrac{1}{2(n+1)}-\mathcal{O}(\dfrac{1}{n^2})$$ and $$n(1+(n+1)\ln\dfrac{n}{n+1})\sim-\dfrac{n}{2(n+1)}-\mathcal{O}(\dfrac1n)\to-\dfrac12$$ as $n\to\infty$.

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My recommendation is to always use asymptotic expansion. But if you want purely elementary means, these inequalities are helpful and can be proven easily: $ \def\lfrac#1#2{{\large\frac{#1}{#2}}} $

  1. $\ln(1+x) \ge \lfrac{2x}{2+x}$ for every real $x \ge 0$.

  2. $\ln(1+x) \le x-\lfrac12x^2+\lfrac13x^3$ for every real $x \ge 0$.

  3. $\ln(1+x) \le \lfrac{2x}{2+x}$ for every real $x \in (-1,0]$.

  4. $\ln(1+x) \ge \lfrac{x}{1+x}+\lfrac12(\lfrac{x}{1+x})^2+\lfrac13(\lfrac{x}{1+x})^3$ for every real $x \in (-1,0]$.

The key is the following fact:

Given any functions $f,g$ on $[0,c]$ where $c \ge 0$, if $f(0) = g(0)$ and $f' \le g'$ then $f \le g$.

Given any functions $f,g$ on $[c,0]$ where $c \le 0$, if $f(0) = g(0)$ and $f' \le g'$ then $f \ge g$.

So to prove (1) and (3) it suffices to prove $\lfrac1{1+x} \ge \lfrac{4}{(2+x)^2}$ for real $x > -1$, which is trivial.

And to prove (2) it suffices to prove $\lfrac1{1+x} \le 1-x+x^2$ for real $x > 0$, which is also trivial.

And (4) immediately falls out from the identity $\ln(1+x) = -\ln(1+\lfrac{-x}{1+x})$ and (2).

Incidentally, the proof for (2) shows that you can easily prove the Taylor series for $\ln(1+x)$ for $x > 0$ by elementary means, since it is an alternating series that converges iff $x < 1$.


Now take any natural $n$.

If $x = -\lfrac1{n+1}$ then $\lfrac{x}{1+x} = -\lfrac1n$ and $\lfrac{2x}{2+x} = -\lfrac{2}{2n+1}$. Thus (3) and (4) immediately give:

  $-\lfrac1{n}+\lfrac1{2n^2}-\lfrac1{3n^3} \le \ln(\lfrac{n}{n+1}) \le -\lfrac{2}{2n+1}$.

Therefore we get:

  $n+(n+1)·(-1+\lfrac1{2n}-\lfrac1{3n^2}) \le n·(1+(n+1)·\ln(\lfrac{n}{n+1})) \le \lfrac{-n}{2n+1}$.

And squeeze theorem does the rest.

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Suppose that $n\le x\le n+1$. Let $u=x-\left(n+\frac12\right)$. Then $-\frac12\le u\le\frac12$ and $$ \begin{align} x(2n+1-x) &=\left(\left(n+\tfrac12\right)+u\right)\left(\left(n+\tfrac12\right)-u\right)\\ &=\left(n+\tfrac12\right)^2-u^2\\ &\in\left[\left(n+\tfrac12\right)^2-\tfrac14,\left(n+\tfrac12\right)^2\right]\tag1 \end{align} $$ Therefore, $$ \begin{align} \log\left(\frac{n+1}n\right) &=\int_n^{n+1}\frac1x\,\mathrm{d}x\tag{2a}\\ &=\frac12\int_n^{n+1}\left(\frac1x+\frac1{2n+1-x}\right)\mathrm{d}x\tag{2b}\\ &=\int_n^{n+1}\frac{n+\frac12}{x(2n+1-x)}\,\mathrm{d}x\tag{2c}\\ &\in\left[\frac1{n+\frac12},\frac1{n+\frac12-\frac1{4n+2}}\right]\tag{2d} \end{align} $$ Explanation:
$\text{(2a)}$: a characterization of $\log$
$\text{(2b)}$: average $\text{(2a)}$ with the substitution $x\mapsto2n+1-x$
$\text{(2c)}$: simplify
$\text{(2d)}$: apply $(1)$

Applying $(2)$, we get $$ \begin{align} n\left(1+(n+1)\log\left(\frac{n}{n+1}\right)\right) &\in\left[n\left(1-\frac{n+1}{n+\frac12-\frac1{4n+2}}\right),n\left(1-\frac{n+1}{n+\frac12}\right)\right]\\ &=\left[n\left(\frac{-\frac12-\frac1{4n+2}}{n+\frac12-\frac1{4n+2}}\right),n\left(\frac{-\frac12}{n+\frac12}\right)\right]\\ &=\left[\color{#C00}{\frac{n}{n+\frac12-\frac1{4n+2}}}\color{#090}{\left(-\frac12-\frac1{4n+2}\right)},\color{#C00}{\frac{n}{n+\frac12}}\color{#090}{\left(-\frac12\right)}\right]\tag3 \end{align} $$ By the Squeeze Theorem, we get $$ \begin{align} \lim_{n\to\infty}n\left(1+(n+1)\log\left(\frac{n}{n+1}\right)\right) &=\color{#C00}{1}\cdot\color{#090}{\left(-\frac12\right)}\\ &=-\frac12\tag4 \end{align} $$

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