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How can I prove $\lim_{n \rightarrow \infty} \ln(n) / \ln(n+1) = 1$ ?

I have looked through all my logarithm rules to find something useful, but the only thing that comes close is $\ln(a/b) = \ln(a)-\ln(b)$.

How can I proceed ?

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Hint

Write

$$\ln(n+1)=\ln\left(n\left(1+\frac1n\right)\right)=\ln n+\ln\left(1+\frac1n\right)$$

Edit We have $$\require{cancel}\frac{\ln n}{\ln(n+1)}=\frac{\ln n}{\ln n+\ln\left(1+\frac1n\right)}=\frac{\cancelto{1}{\ln n}}{\cancel{\ln n}\left(1+\frac{\ln\left(1+\frac1n\right)}{\ln n}\right)}\xrightarrow{n\to\infty}1 $$

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  • $\begingroup$ So we have that their different approach $0$ ? And from this we can conclude that the quotient approach $1$ ? $\endgroup$ – Shuzheng May 1 '14 at 7:38
  • $\begingroup$ You can factor by $\ln n$ and simplify numerator and denominator. Try it! $\endgroup$ – user63181 May 1 '14 at 7:46
  • $\begingroup$ Im not completely sure what you mean. I write $$\ln(n+1) = \ln(n) + \ln(1+1/n) \Rightarrow \ln(n+1)-\ln(n) = \ln(1+1/n) \Rightarrow \ln((n+1)/n) = \ln(1+1/n)$$ ? $\endgroup$ – Shuzheng May 1 '14 at 7:51
  • $\begingroup$ I edited my answer. $\endgroup$ – user63181 May 1 '14 at 7:58
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    $\begingroup$ Ahh, I see it now. Thank you very much for your very clever answer. $\endgroup$ – Shuzheng May 1 '14 at 8:02
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If you apply L'Hopital's Rule, you get

\begin{align}\lim_{n \to \infty} \frac{\ln(n)}{ \ln(n+1)} &= \lim_{n \to \infty} \frac{1/n}{1/(n+1)} \\ &= \lim_{n \to \infty} \frac{n+1}{n} \\ &= 1 \end{align}

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    $\begingroup$ Is this rule also valid when $n$ is a natural number and not a real number ? $\endgroup$ – Shuzheng May 1 '14 at 7:36
  • $\begingroup$ @NicolasLykkeIversen, yes, but I am not sure why you would think might be a problem. $\endgroup$ – IAmNoOne May 1 '14 at 7:38
  • $\begingroup$ @NicolasLykkeIversen, actually are you asking how to find the limit or to actually prove it is equal to $1$? I think I may have misread your intentions in the question. $\endgroup$ – IAmNoOne May 1 '14 at 7:40
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    $\begingroup$ I want to prove it :) $\endgroup$ – Shuzheng May 1 '14 at 7:41
  • $\begingroup$ @NicolasLykkeIversen, actually I think we can still use this result to 'cheat' a bit. By L'Hopital's Rule, your limit is equal to $$\lim_{n \to \infty} \frac{n+1}{n}. $$ Thus we if set $n > \frac{1}{\epsilon}$ for any $\epsilon >0$, then we have established $$ \left |\frac{n+1}{n} - 1 \right | < \epsilon $$ $\endgroup$ – IAmNoOne May 1 '14 at 7:45

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