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With the help of L'Hopital's rule, I found

$$\lim_{n\to \infty}\left(\frac{\log (n-1)}{\log (n)}\right)^n$$

to be equal to $1$.

How can the limit be found without the use of the rule?

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$$A=\left(\frac{\log (n-1)}{\log (n)}\right)^n$$ $$\log(A)=n\log\left(\frac{\log (n-1)}{\log (n)}\right)$$ Now $$\log(n-1)=\log\big(n(1-\frac 1n)\big)=\log(n)+\log\big(1-\frac 1n\big)\approx \log(n)-\frac 1n$$ So $$\frac{\log (n-1)}{\log (n)}\approx 1-\frac 1{n \log(n)}$$ $$\log\left(\frac{\log (n-1)}{\log (n)}\right)\approx -\frac 1{n \log(n)}$$ $$\log(A)\approx -\frac 1{ \log(n)}$$ So, $\lim_{n\to \infty} \log(A)=0$ and then $\lim_{n\to \infty} A=1$

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    $\begingroup$ Personally I'd still prefer $o$ notation to $\approx$ because the latter looks a bit ambiguous as to what type of infinitesimal amount is omitted. $\endgroup$ – Vim Jun 21 '15 at 18:12
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$$\frac{\ln(n-1)}{\ln n}=\frac{\ln n+\ln(1-\frac1n)}{\ln n}=1+\frac{\ln(1-\frac1n)}{\ln n}$$ and by Bernoulli's inequality $$\left(1+\frac{\ln(1-\frac1n)}{\ln n}\right)^n \ge 1+\frac{n\ln(1-\frac1n)}{\ln n}$$ By the Mean Value Theorem, $\ln(1-\frac1n)=-\frac 1n\ln'\xi$ with $1-\frac1n<\xi<1$. Therefore for $n\ge 2$ $$ 1\ge\left(\frac{\ln(n-1)}{\ln n}\right)^n\ge 1-\frac1{2\ln n}$$

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You can write $$\frac{\ln(n-1)}{\ln n} = 1+\frac{\ln(1-\frac{1}{n})}{\ln n} = 1+\frac{-\frac{1}{n}+o(\frac{1}{n})}{\ln n} = 1-\frac{1}{n\ln n} + o\left(\frac{1}{n\ln n}\right) $$ using the Taylor expansion of $\ln(1+x)$ around $0$. Then, $$\begin{align} \left(\frac{\ln(n-1)}{\ln n}\right)^n &= e^{n\ln\left(\frac{\ln(n-1)}{\ln n}\right)} = e^{n\ln\left( 1-\frac{1}{n\ln n} + o\left(\frac{1}{n\ln n}\right)\right)} \\ &= e^{n\left( -\frac{1}{n\ln n} + o\left(\frac{1}{n\ln n}\right)\right)} = e^{ -\frac{1}{\ln n} + o\left(\frac{1}{\ln n}\right)} \\ &\xrightarrow[n\to\infty]{} e^{0} = 1 \end{align}$$ (again, same Taylor expansion). The limit follows from $\frac{1}{\ln n} \xrightarrow[n\to\infty]{} 0$ and the continuity of $\exp$.

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Taylor expansion methods and Hospital's Rule methods are morally the same ideas, even though they seem different at first glance ; the Hospital's rule computes the limit by examining the derivatives of the functions involves and Taylor expansion does the same thing. Here is a technique which is fundamentally different, but as you will see, Taylor expansion really makes things much easier ; this is not a very elementary proof.

One can show using integration that the Euler-Mascheroni constant is well-defined : $$ \gamma \overset{def}= \lim_{n \to \infty} \left( \sum_{i=1}^n \frac 1i - \log (n) \right) \in \, ]0,1[ \quad \Longrightarrow \quad \lim_{n \to \infty} \frac{\sum_{i=1}^n \frac 1i}{\log n} = 1. $$ (this is done by using the definition $\log(x) = \int_1^x \frac 1x \, dx$ and comparing $\int_i^{i+1} \frac 1x \, dx$ with $\frac 1i$ and $\frac 1{i+1}$). Therefore, $$ \lim_{n \to \infty} \left( \frac{\log n}{\log(n-1)} \right)^n = \lim_{n \to \infty} \left( \frac{\left(\sum_{i=1}^{n-1} \frac 1i \right) + \frac 1n}{\sum_{i=1}^{n-1} \frac 1i} \right)^n = \lim_{n \to \infty} \left( 1 + \frac{ \left( \sum_{i=1}^{n-1} \frac 1i \right)^{-1} }n \right)^n = \lim_{n \to \infty} \exp \left( \frac 1{\sum_{i=1}^{n-1} \frac 1i} \right) = \exp(0) = 1. $$ where I used the divergence of the harmonic series and the uniform convergence of the sequence $(1+x/n)^n$ to $e^x$ on the interval $[0,1]$. To prove uniform convergence, you could use Arzela-Ascoli's theorem because on $[0,1]$, the functions $(1+x/n)^n$ are all Lipschitz-continuous with Lipschitz constant $\le e$ : by the $a^n-b^n = (a-b) (\cdots)$ identity, $$ \left| \left( 1 + \frac xn \right)^n - \left( 1 + \frac yn \right)^n \right| \le \left| \frac{x-y}n \right| \left( n \left(1 + \frac 1n \right)^{n-1} \right) \le e|x-y|. $$ The existence of a convergence subsequence via Arzela-Ascoli gives the convergence of the whole sequence since it is pointwise increasing on $[0,1]$ (again a detail to prove, which is a computation usually done when showing that $e^x = \lim_{n \to \infty} (1+x/n)^n$ is well-defined ; with a bit of patience and Bernouilli's inequality you can show that the ratio $(1+x/n)^n/(1+x/(n-1))^{n-1}$ is $\ge 1$ on $[0,\infty[$).

So my point is : use Taylor expansion for those things.

Hope that helps,

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  • $\begingroup$ While a form of Taylor series (namely the one with Peano's reminder) is derived using L'Hospital's Rule, I doubt the L'Hospital's Rule can be derived via Taylor series. $\endgroup$ – Paramanand Singh Jun 22 '15 at 4:35
  • $\begingroup$ @ParamanandSingh : That's precisely via Taylor's theorem and its corollaries that you prove l'Hospital's rule! They're one and the same idea : L'Hospital's rule just says that you should approximate numerator and denominator by their derivatives, and Taylor's theorem does just that. It just requires some thinking, like anything in mathematics : en.wikipedia.org/wiki/… $\endgroup$ – Patrick Da Silva Jun 22 '15 at 9:30
  • $\begingroup$ Wiki link you give does not suggest anywhere that LHR is derived via Taylor series. Rather it is derived via Cauchy's Mean Value Theorem. On the other hand most usual proofs of mean value theorems and also Taylor series (with Lagrange and Cauchy remainders) are given using Rolle's theorem. I still believe that you can't derive LHR via Taylor. If you think otherwise you may produce a proof. $\endgroup$ – Paramanand Singh Jun 22 '15 at 10:04
  • $\begingroup$ @ParamanandSingh : How do you prove that there exists $c < \zeta < x$ such that $\frac{f(x)-f(c)}{x-c} = f'(\zeta)$? This is Cauchy's mean value theorem, I agree. But if you move things around, this equation is $f(x) = f(c) + f'(\zeta) (x-c)$. So Taylor expansion to the first order and Cauchy's mean value theorem are one and the same. This is what I mean when I say these are all the same ideas ; by thinking just a little bit you notice that they are all the same ideas a bit reformulated. $\endgroup$ – Patrick Da Silva Jun 22 '15 at 21:22
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    $\begingroup$ No, Cauchy mvt says that $$\frac{f(b) - f(a)}{g(b) - g(a)} = \frac{f'(c)}{g'(c)}$$ for some $c \in (a, b)$. And this is used to prove LHR. You cant use Taylor of first order to get Cauchy. BTW both Cauchy and Taylor are proved via Rolle's theorem. The situation is like this : if $p \Rightarrow q, p \Rightarrow r$ then we can't say that $q \Leftrightarrow r$. Here $p$ is Rolle, $q$ is LHR and $r$ is Taylor. $\endgroup$ – Paramanand Singh Jun 23 '15 at 2:53
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If $L$ is the desired limit then \begin{align} \log L &= \log\left\{\lim_{n \to \infty}\left(\frac{\log(n - 1)}{\log n}\right)^{n}\right\}\notag\\ &= \lim_{n \to \infty}\log\left(\frac{\log(n - 1)}{\log n}\right)^{n}\text{ (by continuity of log)}\notag\\ &= \lim_{n \to \infty}n\log\left(\frac{\log(n - 1)}{\log n}\right)\notag\\ &= \lim_{n \to \infty}n\log\left(1 + \frac{\log(1 - 1/n)}{\log n}\right)\notag\\ &= \lim_{n \to \infty}n\cdot\frac{\log(1 - 1/n)}{\log n}\cdot\frac{\log n}{\log(1 - 1/n)}\log\left(1 + \frac{\log(1 - 1/n)}{\log n}\right)\notag\\ &= \lim_{n \to \infty}\frac{1}{\log n}\cdot n\log(1 - 1/n)\cdot\lim_{t \to 0}\frac{\log(1 + t)}{t}\text{ (putting }t = \frac{\log(1 - 1/n)}{\log n})\notag\\ &= 0\cdot (-1)\cdot 1 = 0\notag \end{align} Hence $L = 1$.

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As $$\left(\frac{\log(n-1)}{\log n}\right)^n=\left(\frac{\log n+\log\frac{n-1}{n}}{\log n}\right)^n=\left(1+\frac{\log\left(1-\frac{1}{n}\right)}{\log n}\right)^n, $$ by the Binomial Theorem we have $$\lim_{n\to\infty}\left(\frac{\log(n-1)}{\log n}\right)^n=\lim_{n\to\infty} \sum_{k=0}^n{n\choose k}\left(\frac{\log\left(1-\frac{1}{n}\right)}{\log n}\right)^k \\ =\lim_{n\to\infty} \sum_{k=0}^n\frac{n!}{k!(n-k)!}\left(\frac{\log\left(1-\frac{1}{n}\right)}{\log n}\right)^k, $$which, since $\frac{n!}{(n-k)!}\sim n^k$, is the same as $$\lim_{n\to\infty}\sum_{k=0}^n\frac{1}{k!}\left(\frac{n \log\left(1-\frac{1}{n}\right)}{\log n}\right)^k\\=\lim_{n\to\infty} \sum_{k=0}^n\frac{1}{k!}\left(\frac{-1}{\log n}\right)^k=\lim_{n\to\infty}e^{\frac{-1}{\log n}}=1.$$

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No Taylor, no L'Hopital: $\ln (n-1) = \ln n -\int_{n-1}^n dx/x.$ So we have

$$ \ln n - 1/n < \ln (n-1) = \ln n -1/(n+1).$$

Thus

$$1-1/(n\ln n)<\ln (n-1)/\ln n < 1-1/[\ln n(n+1)]$$ $$ \implies (1-1/(n\ln n))^n <(\ln (n-1)/\ln n)^n < (1-1/[(n+1)\ln n])^n.$$

Write the power $n$ as $[(n+1)\ln n]\cdot n/[(n+1)\ln n]$ to see the right hand side $\to (1/e)^0 = 1.$ Same idea for the left hand side. So the limit is $1.$

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