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My friend's trivia league had this math question:

$$\lim_{n \to \infty} \left[\frac{(n+1)^{n + 1}}{n^n} - \frac{n^{n}}{(n-1)^{n-1}} \right]$$

After computing a few values, one could guess the answer is $e$ = 2.718...But how can we prove that is the limit?

Someone offered up a hand-wavy proof like this:

\begin{align} \lim_{n \to \infty} \left[\frac{(n+1)^{n + 1}}{n^n} - \frac{n^{n}}{(n-1)^{n-1}} \right] & = \lim_{n \to \infty} \left[\frac{(n+1)(n+1)^{n}}{n^n} - \frac{n \cdot n^{n-1}}{(n-1)^{n-1}} \right] \\ &= \lim_{n \to \infty} \left[(n+1)\frac{(n+1)^{n}}{n^n} - n\frac{n^{n-1}}{(n-1)^{n-1}} \right] \\ &= \lim_{n \to \infty} \left[(n+1)\left(1 + \frac{1}{n} \right)^n - n\left(\frac{n - 1 + 1}{n-1} \right)^{{n-1}} \right] \\ &= \lim_{n \to \infty} \left[(n+1)\left(1 + \frac{1}{n} \right)^n - n\left(1 + \frac{1}{n-1} \right)^{n-1} \right] \\ &= \lim_{n \to \infty} \left[(n+1)e - n \cdot e \right] \\ &= \lim_{n \to \infty} e \\ &= e \end{align}

The part about substituting $e$ is hand-wavy since technically this is an indeterminate form of $\infty - \infty$. And using $e$ as as upper bound did not lead me to an easy proof either.

Is there a way to rigorously prove the limit? I tried a few approaches: (a) sandwiching the limit--I could prove $e$ was a lower bound, but I could not find a suitable upper bound converging to $e$, (b) using L'Hopital's rule with no luck, (c) using the mean value theorem--but that also got complicated.

So this is a pretty tough problem to ask at trivia! Is there a way to prove this limit formally?

Sources

Trivia question: http://learnedleague.com/question.php?72&16&4

Thread on trivia: http://learnedleague.com/viewtopic.php?f=10&t=7961&hilit=euler

Hand-wavy proof: http://imgur.com/rIXghhw

Idea for mean value theorem: http://www.pharout.com/trickylimitproblem.pdf

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    $\begingroup$ +1 for recognizing the hand-waviness of the alleged "proof". $\endgroup$ – dxiv Mar 28 '17 at 3:48
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Mean value theorem approach: Define

$$f(x) = \frac{(x+1)^{x+1}}{x^x}.$$

Logarithmic differentiation shows

$$\tag 1 f'(x) = \frac{(x+1)^{x+1}}{x^x}\ln (1+1/x) = (1+1/x)^x(x+1)\ln (1+1/x).$$

We want the limit of $f(x)-f(x-1).$ By the MVT and $(1),$ this difference equals

$$(1+1/c)^c(c+1)\ln (1+1/c)\cdot 1$$

for some $c=c_x \in (x-1,x).$ As $x\to \infty,$ $c_x\to \infty.$ Now we know $(1+1/c)^c \to e,$ and verifying $(c+1)\ln (1+1/c) \to 1$ is standard. The desired limit is therefore $e.$

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  • $\begingroup$ Thanks for filling in the details. In case others need more detail, the limit of (c + 1)ln(1 + 1/c) is also an indeterminate form. It can be converted to 0/0 by re-writing as ln(1 + 1/c)/(1/(c+1)). Then use L'Hopitals rule--the derivative is ln(1 + 1/c) - 1/c. This goes to 1 as c goes to infinity. Then we can use the product rule of limits for (1 + 1/c)^c and the rest of the function. $\endgroup$ – Presh Mar 29 '17 at 21:48
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Here, we present a rigorous development that is based on asymptotic expansions. Proceeding, we have

$$\begin{align} \frac{(n+1)^{n + 1}}{n^n} - \frac{n^{n}}{(n-1)^{n-1}}&=(n+1)\left(1+\frac{1}{n}\right)^n-n\frac{1}{\left(1-\frac1n\right)^{n-1}}\\\\ &=(n+1)e^{n\log\left(1+\frac{1}{n}\right)}-ne^{-(n-1)\log\left(1-\frac{1}{n}\right)}\\\\ &=(n+1)e^{1-\frac{1}{2n}+O\left(\frac1{n^2}\right)}-ne^{1-\frac1{2n}+O\left(\frac{1}{n^2}\right)}\\\\ &=e\left(n+\frac12+O\left(\frac{1}{n}\right)\right)-e\left(n-\frac12+O\left(\frac1n\right)\right)\\\\ &=e+O\left(\frac1n\right) \end{align}$$

Taking the limit as $n\to \infty$, we have

$$\lim_{n\to \infty}\left(\frac{(n+1)^{n + 1}}{n^n} - \frac{n^{n}}{(n-1)^{n-1}}\right)=e$$

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  • $\begingroup$ Thanks! I like this solution, though I will have to read up on asymptotic expansions. I'll take suggestions, here are a couple of resources I found from a first search: math.ucdavis.edu/~hunter/m204/ch2.pdf and web2.clarkson.edu/projects/subramanian/ch561/notes/… $\endgroup$ – Presh Mar 29 '17 at 21:52
  • $\begingroup$ You're welcome! My pleasure. -Mark $\endgroup$ – Mark Viola Mar 29 '17 at 21:54
  • $\begingroup$ The expansions herein are from Taylor's Theorem. $\endgroup$ – Mark Viola Mar 29 '17 at 21:55
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    $\begingroup$ I solved in the same way! (at the dupe) $\endgroup$ – user Dec 8 '18 at 2:21
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We can see that $$A = \frac{(n + 1)^{n + 1}}{n^{n}} = \exp((n + 1)\log (n + 1) - n \log n) = \exp(\log n + (n + 1)\log(1 + 1/n))$$ and similarly $$B = \frac{n^{n}}{(n - 1)^{n - 1}} = \exp(\log n - (n - 1)\log(1 - 1/n))$$ and subtracting we get $$A - B = e^{a} - e^{b} = e^{b}(e^{a - b} - 1) = e^{b}\cdot\frac{e^{a - b} - 1}{a - b}\cdot (a - b)$$ where \begin{align} a - b &= (n + 1)\log(1 + 1/n) + (n - 1)\log(1 - 1/n)\notag\\ & = \log(1 + 1/n) - \log(1 - 1/n) + n\log(1 - 1/n^{2})\notag \end{align} which shows that $a - b \to 0$ and therefore $(e^{a - b} - 1)/(a - b) \to 1$. Moreover we know that $B/n = e^{b}/n \to e$. The desired limit is given by $$\lim_{n \to \infty}e^{b}(a - b) = \lim_{n\to\infty}\frac{B}{n}\cdot n(a - b)$$ and clearly $n(a - b) \to 1$ so that the overall limit is $e$.


In the above we have used the following facts $$e = \lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^{n}, \lim_{x \to 0}\frac{\log(1 + x)}{x} = 1$$

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    $\begingroup$ Great solution here! I was just inquiring how can deal with that by standard limits and if it was possible! $\endgroup$ – user Dec 8 '18 at 2:20
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Almost as in the same spirit as in Dr. MV's answer, consider $$A_n=\frac{(n+1)^{n + 1}}{n^n}\implies \log(A_n)=(n+1)\log(n+1)-n\log(n)$$ Now, using Taylor $$\log(A_n)=1+\log \left({n}\right)+\frac{1}{2 n}-\frac{1}{6 n^2}+\frac{1}{12 n^3}+O\left(\frac{1}{n^4}\right)$$ Taylor again $$A_n=e^{\log(A_n)}=e n+\frac{e}{2}-\frac{e}{24 n}+\frac{e}{48 n^2}+O\left(\frac{1}{n^3}\right)$$ Doing the same for the second piece $A_{n-1}$, we then have $$\frac{(n+1)^{n + 1}}{n^n} - \frac{n^{n}}{(n-1)^{n-1}}=e+\frac e{24n^2}+O\left(\frac{1}{n^3}\right)$$ which shows the limit and how it is approached.

Just for the fun, use $n=10$ and notice that the difference between the exact and approximated value is $\approx 4.7\times 10^{-6}$.

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