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Tried this question here How to calculate $\lim\limits_{n \to \infty} \left( \frac{\ln(n+1)^{n+1}}{\ln n^n} \right)^n$? and was curious about the result. The answer according to Wolfram Alpha is $e$, so I wanted to try it.

$\lim\limits_{n \to \infty} \left( \frac{\ln((n+1)^{n+1})}{\ln (n^n)} \right)^n$

$\lim\limits_{n \to \infty} \left( \frac{(n+1)\ln(n+1)}{n\ln (n)} \right)^n$

$\lim\limits_{n \to \infty} \left( \frac{\ln(n+1)}{\ln(n)}\left(1 + \frac{1}{n}\right) \right)^n$

This is similar to the typical definition $\lim\limits_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e$ but it has the extra log factors.

How come these two happen to be equivalent? Is it valid to apply L'Hospital's Rule to the logs even though they're inside the $()^n$? Or can it be applied to just part of the function and not the other half? What's the correct way to handle this extra log multiplier?

For instance:

$\lim\limits_{n \to \infty}\frac{\ln(n+1)}{\ln(n)} = \lim\limits_{n \to \infty}\frac{\frac{d}{dn}\ln(n+1)}{\frac{d}{dn}\ln(n)} = \lim\limits_{n \to \infty}\frac{n}{1+n} = \lim\limits_{n \to \infty}\frac{1}{1/n+1} = 1$

but I don't think we can necessarily analyze this "separately" from the main result; I think they must be taken together somehow. I also considered squeeze theorem but couldn't think of another function approaching $e$ from the other side.

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  • $\begingroup$ The log fraction goes to 1 as n goes to infinity, no? $\endgroup$ – pie314271 Jan 21 '17 at 13:58
  • $\begingroup$ Yes, but I want a more formal proof; in practice if I get in the habit of trying to eyeball things, I get it wrong $\endgroup$ – Jay Smith Jan 21 '17 at 13:58
  • $\begingroup$ The intuition says that $\ln(n+1)/\ln(n)\approx 1$ as $n$ gets large, but of course this does not constitute a rigorous proof. $\endgroup$ – Eff Jan 21 '17 at 13:59
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    $\begingroup$ @pie314271 (and Eff too): that is an invalid argument. The $(1+\frac{1}{n})$ part tends to $1$ too, but still the limit is not $1$. $\endgroup$ – TonyK Jan 21 '17 at 14:00
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    $\begingroup$ I don't think I showed that $\endgroup$ – Jay Smith Jan 21 '17 at 14:06
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With a Taylor expansion-based argument:

When $n\to\infty$, we get $$ \frac{\ln(n+1)}{\ln n}= \frac{\ln n+\ln\left(1+\frac{1}{n}\right)}{\ln n} = 1+ \frac{\ln\left(1+\frac{1}{n}\right)}{\ln n} = 1 + \frac{1}{n\ln n} + o\left(\frac{1}{n\ln n}\right) \tag{1} $$ (using that $\ln(1+x)=x+o(x)$ when $x\to0$) so that $$\begin{align} \frac{\ln(n+1)}{\ln n}\left(1+\frac{1}{n}\right) &= \left(1 + \frac{1}{n\ln n} + o\left(\frac{1}{n\ln n}\right)\right)\left(1+\frac{1}{n}\right) = 1+\frac{1}{n}+\frac{1}{n\ln n} + o\left(\frac{1}{n\ln n}\right)\\ &= 1+\frac{1}{n}+o\left(\frac{1}{n}\right) \tag{2} \end{align}$$ and from (2) and the same Taylor expansion of $\ln(1+x)$ at $0$ we get $$\begin{align} \left(\frac{\ln(n+1)}{\ln n}\left(1+\frac{1}{n}\right)\right)^{n} &= e^{n\ln \left(\frac{\ln(n+1)}{\ln n}\left(1+\frac{1}{n}\right)\right)} = e^{n\ln \left(1+\frac{1}{n}+o\left(\frac{1}{n}\right)\right)} = e^{n\left(\frac{1}{n}+o\left(\frac{1}{n}\right)\right)} = e^{1+o\left(1\right)} \\&\xrightarrow[n\to\infty]{} e^1 = e \end{align}$$ as claimed.

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  • $\begingroup$ Those are slick transformations / rearrangements but I don't think I would have ever seen those on my own; are they typical strategies or is there a logical process behind the rearrangement at each step? $\endgroup$ – Jay Smith Jan 21 '17 at 14:17
  • $\begingroup$ Also how do you go from $1+\frac{1}{n}+\frac{1}{n\ln n} + o\left(\frac{1}{n\ln n}\right)$ to $1+\frac{1}{n}+o\left(\frac{1}{n}\right)$? What does $o$ mean here exactly? $\endgroup$ – Jay Smith Jan 21 '17 at 14:19
  • $\begingroup$ @JaySmith I am heavily biased on using Taylor expansions whenever possible (as it's a systematic way to tackle limits, and works almost every time). With that in mind, you want to make "standard" quantities appear, with something tending to $0$: the thing that is mainly the issue here is the ratio of logarithms, so the first thing to do is massage it. $$\ln(n+1) = \ln n + \ln\left(1+\frac{1}{n}\right)$$ is a good thing to do to achieve our goals, because of that. After this step, it's pretty much on autopilot. $\endgroup$ – Clement C. Jan 21 '17 at 14:20
  • $\begingroup$ The second (and last) standard and very useful step is to write the quantity of interest $f(n)^{g(n)}$ in its exponential form $\exp(g(n)\ln f(n))$: by continuity of $\exp$, one then only has to care about the exponent $g(n)\ln f(n)$, and this does almost always make things much clearer. $\endgroup$ – Clement C. Jan 21 '17 at 14:21
  • $\begingroup$ @JaySmith $o(\cdot)$ ("little o") is the Landau notation. One writes $f(n)=o(g(n))$ (at some specified point, here when $n\to \infty$) if, basically, $\frac{f(n))}{g(n)}\xrightarrow[n\to\infty]{} 0$. In our case, we have $\frac{\frac{1}{n\ln n}+o\left(\frac{1}{n\ln n}\right)}{\frac{1}{n}} = \frac{1}{\ln n} + o\left(\frac{1}{\ln n}\right) \xrightarrow[n\to\infty]{} 0$. $\endgroup$ – Clement C. Jan 21 '17 at 14:23
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Use my comment in the question mentioned to use that if $a_n\to a$ then $$\left(1+\frac{a_n}{n}\right)^n \to e^a$$ in this case $$a_n=n\frac{\ln(n+1)-\ln n}{\ln n}=\frac{1}{\ln n}\ln \left(1+\frac{1}{n}\right)^n\to 0$$ and thus $$\left(\frac{\ln (n+1)}{\ln n}\right)^n\to 1$$

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