Why is $$\lim_{n\to\infty}\left(\frac{\log(n+1)}{\log n}\right)^{n}=1$$
Can I compute the part inside the square bracket first?
Thanks for helping.
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You cannot compute the part inside the square brackets first, because $1^\infty$ is indeterminate. – Eckhard Mar 21 '13 at 11:12
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Is the limit equals to 1 then? – Xiaobin Zhu Mar 21 '13 at 11:16
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it equals 1 yes – Dominic Michaelis Mar 21 '13 at 11:18
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The standard way to compute limits of this kind is to use the well-known limit: $$\lim_{x\to\infty}\left(1+\frac1x\right)^{x}=e=\lim_{x\to0}(1+x)^{\frac1x}$$ In your case, rewrite the limit as: $$\lim_{n\to\infty}\left(\frac{\log(n+1)}{\log n}\right)^{n}=\lim_{n\to\infty}\left[\left(1+\frac{\log(n+1)-\log n}{\log n}\right)^{\frac{\log n}{\log(n+1)-\log{n}}}\right]^{\frac{\log(n+1)-\log n}{\log n}n}$$ Now use the aforementioned limit and compute the $\lim_{n\to\infty}\frac{\log(n+1)-\log n}{\log n}n$ separately.
Dennis Gulko
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Take $\log$s and use $\log (x+y) \leq \log x + y/x$ (which is true by MVT) twice. Note
\begin{align*} \log\left(\frac{\log(n+1)}{\log n}\right)^n &= n \log\log(n+1) - n \log\log n \\ &\leq n\log(\log n + 1/n) - n \log \log n \\ &\leq n(\log\log n + 1/(n \log n)) - n\log\log n\\ &=1/\log n \longrightarrow 0 \end{align*}
Sean Eberhard
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Using the equivalence $$\log{\left(1+\dfrac{1}{n}\right)}\underset{n\to\infty}{\sim} \dfrac{1}{n}$$ we have $$\dfrac{\log(n+1)}{\log n}=\dfrac{\log \left(n\left(1+\dfrac{1}{n}\right)\right)}{\log n}=\dfrac{\log {n}+\log{\left(1+\dfrac{1}{n}\right)}}{\log n}=\\ =1+\dfrac{\log{\left(1+\dfrac{1}{n}\right)}}{\log n}\underset{n\to\infty}{\sim} 1+\dfrac{1}{n\log n}.$$ Then $$\lim_{n\to\infty}\left(\frac{\log(n+1)}{\log n}\right)^{n}=\lim_{n\to\infty}\left(1+\dfrac{1}{n\log n}\right)^{n}=\lim_{n\to\infty}{e^{\frac{1}{\log{n}}}}=1.$$
M. Strochyk
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There is something wrong with this: $\log{\left(1+\dfrac{1}{n}\right)}\underset{n\to\infty}{\sim} 1+\dfrac{1}{n}$ because $\log{1} = 0$. I can see $\underset{n\to\infty}{\sim} \dfrac{1}{n}$ but not as it is stated above. – John Nicholson Mar 30 '15 at 04:44
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@John Nicholson: Thanks, you are right. Edited. It was a mechanical mistake; the next lines are correct. – M. Strochyk Mar 30 '15 at 14:58
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We'll write the function in $e^{\log}$ form$$\Big(\dfrac{\log(n+1)}{\log n}\Big)^n=\exp(n\log(\dfrac{\log(n+1)}{\log(n)}))=\exp(n \log(1+\dfrac{1}{n\log n }+o(\frac{1}{n^2})))$$$$=\exp(\frac{1}{\log(n)}+o(\frac{1}{n}))$$ Taking the limit we get 1.The 2nd step is by $\log(1+x)=x +o(x^2)$ for x close to 0.
Ishan Banerjee
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