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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

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5  
What the heck is "guz"? –  user2357112 Jun 8 at 0:07

4 Answers 4

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.

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

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I think you meant "as $n\to \infty$" –  chubakueno Jun 7 at 16:17
    
@chubakueno yes he meant "as $n→∞$" –  Educ Jun 7 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) $\lim_{n \to \infty}\frac{\pi \log (a^{2n}-1)}{n} = \frac{\pi (\log (a^{2n}) + \log (1-\frac{1}{a^{2n}}))}{n} \sim 2 \pi \log a -\frac{\pi}{n a^{2n}} \to_{n} 2 \pi \log a$

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yes $\log (1 -x ) \sim -x $ as $x \to 0$ –  Alex Jun 7 at 17:05

$${(\alpha-1)^2(\alpha^{2n}-1)\over(\alpha+1)(\alpha-1)}=\alpha^{2n}(1-\alpha^{-2n}){\alpha-1\over\alpha+1}$$

so

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

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 $\ln(1-\alpha^{-2n})\to\ln(1-0)=\ln1=0$.

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That's awesome Thanks –  Educ Jun 7 at 17:02

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