Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For $\alpha>1$ how can I show that $\lim\limits_{n\to \infty}\dfrac{\pi}{n}\left( \ln \left(\dfrac{\left(α-1\right)^{2}\left(α^{2n}-1\right)}{(α+1)(\alpha-1)}\right)\right)=2\pi\ln(\alpha)$

Indeed, $\lim\limits_{n\to \infty}\dfrac{\pi}{n}\left( \ln \left(\dfrac{(α-1)^{2}(α^{2n}-1)}{(α+1)(\alpha-1)}\right)\right)=\lim\limits_{n\to \infty}\left(\dfrac{\pi \ln \left(\dfrac{(α-1)(α^{2n}-1)}{α+1}\right)}{n}\right)=$

since $\alpha>1$ then $\lim\limits_{n\to\infty\:}\alpha^{2n}-1=+\infty$

Any help will be appreciated

share|cite|improve this question
What the heck is "guz"? – user2357112 Jun 8 '14 at 0:07
up vote 1 down vote accepted

Use L'Hospital for $$\dfrac{ \ln \left(\dfrac{\left(α-1\right)\left(α^{2n}-1\right)}{α+1}\right)}{n}=\dfrac{ \ln \left(\dfrac{\left(α-1\right)\left(e^{2n\log α}-1\right)}{α+1}\right)}{n}$$ After very few simplifications, the derivative of the numerator (with respect to $n$) is just $$\frac{2 α^{2 n} \log (α)}{α^{2 n}-1}$$

I am sure that you can take from here and conclude.

share|cite|improve this answer
Not at all.$\lim_{n\to+\infty} \frac{2 a^{2 n} \log (a)}{a^{2 n}-1}=\lim_{n\to+\infty} \frac{ a^{2 n} 2\log (a)}{a^{2 n}}=2~\log(a)$ – Claude Leibovici Jun 7 '14 at 16:30

Another way is just to use the laws of logarithms, $$\ln\left(\frac{(\alpha-1)(\alpha^{2n}-1)}{\alpha+1}\right) =\ln (\alpha-1)+\ln (\alpha^{2n}-1)-\ln(\alpha+1)$$ Now $\frac{\ln(\alpha-1)}{n}\to 0$ as $n\to 0$ same for similar terms. For the remaining term, $$\ln (\alpha^{2n}-1)=\ln (\frac{\alpha^{2n}-1}{\alpha^{2n}})+\ln (\alpha^{2n})$$ and $\frac{\alpha^{2n}-1}{\alpha^{2n}}\to 1$ so $$\frac{1}{n}\ln (\frac{\alpha^{2n}-1}{\alpha^{2n}})\to 0$$

What remains is $$\frac{\pi}{n}\ln (\alpha^{2n})=2\pi\ln (\alpha)$$

share|cite|improve this answer
I think you meant "as $n\to \infty$" – chubakueno Jun 7 '14 at 16:17
@chubakueno yes he meant "as $n→∞$" – Educ Jun 7 '14 at 16:44

1) Use $\log(xy) = \log x + \log y$ and $\log \frac{x}{y} = \log x - \log y$ to simplify everything except the $\log (a^{2n}-1)$ term; all of them tend to 0

2) $\lim_{n \to \infty}\frac{\pi \log (a^{2n}-1)}{n} = \frac{\pi (\log (a^{2n}) + \log (1-\frac{1}{a^{2n}}))}{n} \sim 2 \pi \log a -\frac{\pi}{n a^{2n}} \to_{n} 2 \pi \log a$

share|cite|improve this answer
yes $\log (1 -x ) \sim -x $ as $x \to 0$ – Alex Jun 7 '14 at 17:05



$${\pi\over n}\ln\left({(\alpha-1)^2(\alpha^{2n}-1)\over(\alpha+1)(\alpha-1)}\right)=2\pi\ln(\alpha)+{\pi\over n}\left(\ln(1-\alpha^{-2n})+\ln(\alpha-1)-\ln(\alpha+1)\right)$$

The limit of the final three terms as $n\to\infty$ is clearly $0$. In particular, you don't even need the $1/n$ for the first term, since $\alpha\gt1$ implies $\ln(1-\alpha^{-2n})\to\ln(1-0)=\ln1=0$.

share|cite|improve this answer
That's awesome Thanks – Educ Jun 7 '14 at 17:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.