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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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    The log fraction goes to 1 as n goes to infinity, no? – 2017-01-21
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    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 – 2017-01-21
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    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. – 2017-01-21
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    @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 $1$. – 2017-01-21
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    You need to show that $\left(\frac{\ln(n+1)}{\ln n}\right)^n$ tends to $1$. Then you can use the fact that if $\lim a_n =a$ and $\lim b_n=b$, then $\lim a_nb_n$ exists and equals $ab$. – 2017-01-21
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    Oh, you just did that! Then you are finished :-) – 2017-01-21
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    I don't think I showed that – 2017-01-21
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    @TonyK One can still be intuitive. $(1+\frac 1n)^{an}$ shows that if the base goes to $1$ and the exponent goes to $\infty$ in an inversely proportional manner, then the limit is between $1$ and $\infty$. If one converges faster than inversely proportional to the other, then the limit is $1$ or $\infty$, depending on which is faster. Of course, this is not proof either, but it's closer to true. – 2017-01-21
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    the limit of $$\left(\frac{\ln(n+1)}{\ln(n)}\right)^n=1$$ for $n$ tends to infinity – 2017-01-21
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    @JaySmith: What didn't you show (in your opinion)? – 2017-01-21
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    I showed that $\frac{\ln(n+1)}{\ln(n)}$ goes to $1$, not $\left(\frac{\ln(n+1)}{\ln(n)}\right)^n$ – 2017-01-21
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    So you did. My mistake. – 2017-01-21

2 Answers 2

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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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    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? – 2017-01-21
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    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? – 2017-01-21
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    @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. – 2017-01-21
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    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. – 2017-01-21
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    @JaySmith $o(\cdot)$ ("little o") is the [Landau notation](https://en.wikipedia.org/wiki/Big_O_notation#Related_asymptotic_notations). 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 $\frac{\frac{1}{n\ln n}+o\left(\frac{1}{n\ln n}\right)}{\frac{1}{n}} = \frac{1}{\ln n} + o\left(\frac{1}{\ln n}\right) \xrightarrow[n\to\infty]{} 0$. – 2017-01-21
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    I am also a little confused where the actual Taylor expansion is being used; if I Google it I see $\ln(x + 1) = x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots$ as the "Taylor expansion" – 2017-01-21
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    @JaySmith Yes -- the Taylor expansion is an infinite series, but here we do a Taylor expansion to finite order. This is also called [Taylor's theorem, with the Peano form of the remainder.](https://en.wikipedia.org/wiki/Taylor's_theorem#Statement_of_the_theorem) – 2017-01-21
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    I'm with you on that rearrangement but still don't know why we say that $\frac{1}{n\ln n} + o\left(\frac{1}{n\ln n}\right)$ is the same as $o\left(\frac{1}{n}\right)$ – 2017-01-21
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    Using the definition: $$\frac{1}{n\ln n} + o\left(\frac{1}{n\ln n}\right)$$ stands for "$\frac{1}{n\ln n} + f(n)$ for some function $f$ such that $\frac{f(n)}{\frac{1}{n\ln n}} \xrightarrow[n\to\infty]{} 0$." Now, we have $$\frac{\frac{1}{n\ln n}+f(n)}{\frac{1}{n}} = \frac{1}{\ln n} + \frac{1}{\ln n}\cdot \frac{f(n)}{\frac{1}{n\ln n}} \xrightarrow[n\to\infty]{} 0+0\cdot 0 = 0$$ which means by definition that $$\frac{1}{n\ln n}+f(n) = o\left(\frac{1}{n}\right)$$ @JaySmith – 2017-01-21
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    The above is a formal way to prove it, byut after a while you'll just use the fact that arithmetic (sum, multiplication, division) with $o(\cdot)$'s pretty much works like one would expect. – 2017-01-21
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    Thanks for the explanation – 2017-01-21
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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 $$\left(\frac{\ln (n+1)}{\ln n}\right)^n\to 1$$