Limit of $n\ln\left(\frac{n-1}{n+1}\right)$

Question

I'm having trouble with this limit:

$$\lim_{n\to ∞}n\ln\left(\frac{n-1}{n+1}\right)$$

It's supposed to be solvable without using l'Hospital's rule. I'm guessing it's a case for the squeeze theorem, but I'm not exactly sure. Any advice?

score 3 · Answer 1 · answered Dec 03 '15 at 22:42

3

You can write $$ \frac{n-1}{n+1}=\frac{n+1-2}{n+1}=1-\frac{2}{n+1} $$ so you can also write your limit as $$ \lim_{n\to\infty}\frac{n}{n+1}(n+1)\ln\left(1-\frac{2}{n+1}\right) = \lim_{n\to\infty}\frac{n}{n+1} \ln\biggl(\biggl(1-\frac{2}{n+1}\biggr)^{n+1}\biggr) $$ Now you should know that $$ \lim_{n\to\infty}\biggl(1+\frac{a}{n}\biggr)^n=e^a $$

answered Dec 03 '15 at 22:42

egreg

238,574

This was helpful too – Jack Krasowsky Dec 03 '15 at 23:07

Kay K. · Accepted Answer · 2015-12-03T22:52:30.810

2

\begin{align} &\lim_{n\to \infty}n\ln\left(\frac{n-1}{n+1}\right)=\lim_{n\to \infty}\ln\left(\frac{n+1-2}{n+1}\right)^n\\ &=\lim_{n\to \infty}\ln\left(1-\frac{2}{n+1}\right)^n\\ &=\lim_{n\to \infty}\ln\left(1-\frac{2}{n+1}\right)^{\frac{-(n+1)}{2}\cdot\frac{-2n}{n+1}}\\ &=\lim_{n\to \infty}\ln\left(\left(1-\frac{2}{n+1}\right)^{\frac{-(n+1)}{2}}\right)^{\frac{-2n}{n+1}}\\ &=\lim_{n\to \infty}\ln \left(e^{\frac{-2n}{n+1}}\right)=\lim_{n\to \infty}\frac{-2n}{n+1}=-2 \end{align}

edited Dec 03 '15 at 22:52

answered Dec 03 '15 at 22:48

Kay K.

9,931

How did you come up with the exponent in step 3? – Jack Krasowsky Dec 03 '15 at 22:50
@JackKrasowsky I added one more line to elaborate it. – Kay K. Dec 03 '15 at 22:52
I meant more like what happened between steps 2 and 3 – Jack Krasowsky Dec 03 '15 at 22:57
@JackKrasowsky because $\frac{-(n+1)}{2}\cdot\frac{-2n}{n+1}=n$. – Kay K. Dec 03 '15 at 22:59
Of course, but was there any special reason for writing it this way? Does it have a relationship with the argument of the log function? – Jack Krasowsky Dec 03 '15 at 23:03
@JackKrasowsky I got your question now. Why I did like that was I tried to make the form of $\lim_{x\to0}(1+x)^{\frac1x}$ that is the definition for $e$. – Kay K. Dec 03 '15 at 23:05
Right, I see it now! Thanks – Jack Krasowsky Dec 03 '15 at 23:06
You should probably put the entire argument to $\ln$ inside parenthesis to make it clear that it is the argument that is raised to the power of $n$ and not the result of $\ln$. This got me confused for a while. – HelloGoodbye Jan 20 '16 at 12:32

score 1 · Answer 3 · answered Dec 03 '15 at 22:53

1

With equivalents: $\;\ln(1-u)\sim_0 -u$, hence $$n\ln\Bigl(\frac{n-1}{n+1}\Bigr)=n\ln\Bigl(1-\frac{2}{n+1}\Bigr)\sim_\infty-\frac{2n}{n+1}\sim_\infty -2.$$

answered Dec 03 '15 at 22:53

Bernard

175,478

This is too advanced for me I'm afraid – Jack Krasowsky Dec 03 '15 at 23:11
Didn't you see some asymptotic analysis and Taylor's formula? – Bernard Dec 03 '15 at 23:25

score 1 · Answer 4 · edited Apr 13 '17 at 12:20

1

In THIS ANSWER, I showed using standard tools that don't rely on derivatives that, for $z>-1$, the logarithm function satisfies the inequalities

$$\frac{z}{z+1}\le \log (1+z)\le z$$

Now, we note that $\log \left(\frac{n-1}{n+1}\right)=\log \left(1-\frac{2}{n+1}\right)$. Therefore, we have for $n>1$

$$-\frac{2n}{(n-1)}\le n\log \left(1-\frac{2}{n+1}\right)\le -\frac{2n}{n+1}$$

Using the Squeeze Theorem, we obtain the limit

$$\lim_{n\to \infty}n\,\log \left(\frac{n-1}{n+1}\right)=-2$$

edited Apr 13 '17 at 12:20

Community

1

answered Dec 03 '15 at 23:13

Mark Viola

179,405

Please let me know how I can improve my answer. I really want to give you the best answer I can. - Mark – Mark Viola Dec 03 '15 at 23:17

score 0 · Answer 5 · answered Dec 03 '15 at 22:39

0

Notice, let $\frac{1}{n}=t$, hence $$\lim_{n\to \infty}n\ln\left(\frac{n-1}{n+1}\right)$$ $$=\lim_{t\to 0}\frac{1}{t}\ln\left(\frac{\frac{1}{t}-1}{\frac{1}{t}+1}\right)$$ $$=\lim_{t\to 0}\frac{\ln\left(\frac{1-t}{1+t}\right)}{t}$$ $$=\lim_{t\to 0}\frac{\ln\left(1-t\right)-\ln\left(1+t\right)}{t}$$ Now, using Taylor's series, $$=\lim_{t\to 0}\frac{-\left(t+\frac{t^2}{2}+O(t^2)\right)-\left(t-\frac{t^2}{2}+O(t^2)\right)}{t}$$ $$=\lim_{t\to 0}\left(-\left(1+\frac{t}{2}+O(t)\right)-\left(1-\frac{t}{2}+O(t)\right)\right)$$ $$=-\left(1+0\right)-\left(1-0\right)=\color{red}{-2}$$

answered Dec 03 '15 at 22:39

Harish Chandra Rajpoot

37,450

What's the O thing supposed to be? – Jack Krasowsky Dec 03 '15 at 22:41
abbreviation for higher power terms as the series has infinite terms – Harish Chandra Rajpoot Dec 03 '15 at 22:44
Alright, thanks – Jack Krasowsky Dec 03 '15 at 22:44
Why do you write $\ln\left(1-t\right)-\ln\left(1+t\right)$ as $ -\left(t+\frac{t^2}{2}+O(t^2)\right)-\left(t-\frac{t^2}{2}+O(t^2)\right)$ and not just as $-\left(t+O(t^2)\right)-\left(t+O(t^2)\right)$? The former is not incorrect and it will lead to the correct answer but it does not make much sense to me. The $\frac{t^2}{2}$ terms have no effect since you have the $O(t^2)$ terms. Besides the Taylor expansion of $\ln(1+t)$ around $t=0$ is $t-t^2+...$, not $t-\frac{t^2}{2}+...$. – HelloGoodbye Jan 20 '16 at 12:49

score 0 · Answer 6 · answered Dec 04 '15 at 10:41

Quite simply one needs to use the standard limit formula $$\lim_{x \to 0}\frac{\log(1 + x)}{x} = 1\tag{1}$$ We have \begin{align} \lim_{n \to \infty}n\log\left(\frac{n - 1}{n + 1}\right) &= \lim_{n \to \infty}n\log\left(1 - \frac{2}{n + 1}\right)\notag\\ &= \lim_{n \to \infty}n\cdot\frac{-2}{n + 1}\cdot\dfrac{\log\left(1 - \dfrac{2}{n + 1}\right)}{-\dfrac{2}{n + 1}}\notag\\ &= -2\lim_{n \to \infty}\frac{n}{n + 1}\cdot\lim_{x \to 0}\frac{\log(1 + x)}{x}\text{ (putting }x = -\frac{2}{n + 1})\notag\\ &= -2\cdot 1\cdot 1 = -2\notag \end{align}

Limit of $n\ln\left(\frac{n-1}{n+1}\right)$

6 Answers6