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These days I saw a really interesting limit as I was reading more information on Napier's constant here : http://mathworld.wolfram.com/e.html. It seems a pretty young limit since it appears under the name and year "Brothers and Knox 1998". It's also new for me and I'd like to know more approaching ways for it.

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

A possible way to go is to use the first converse of the Stolz–Cesàro theorem, but since $\lim_{n\to\infty}\frac{b_{n}}{b_{n+1}}=1$ we can at most check the given result because the theorem may work
or not in this case. For the case when $\lim_{n\to\infty}\frac{b_{n}}{b_{n+1}}=1$ it is required more research in
order to be sure that we can safely apply the first converse theorem. I'll make an update
as soon as things are clarified such that I may turn it into a rigorous proof.

$\lim_{n\to\infty} \frac{{(n+1)}^{n+1}}{n^n} - \frac{{n}^{n}}{{(n-1)}^{n-1}}=$ $\lim_{n\to\infty} f(n+1) - f(n)=$ By the first converse of the Stolz–Cesàro theorem we have $\lim_{n\to\infty} \frac{f(n+1) - f(n)}{n+1-n}=$ $\lim_{n\to\infty} \frac{f(n)}{n}=$ $\lim_{n\to\infty} \frac{n^n}{(n-1)^{n-1}}\cdot \frac{1}{n}=$ $\lim_{n\to\infty} \frac{n^{n-1}}{(n-1)^{n-1}}=\left(1+\frac{1}{n-1}\right)^{n-1}=e.$

What else can we do here? Thanks.

  • 0
    At $a$ny rate you should not say that yours is a proof... It is misleading for unexperienced users.2012-09-11

2 Answers 2

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$\eqalign{(n+1)^{n+1} = \exp\left((n+1) \ln(n+1)\right) &= \exp\left(((n+1) (\ln n + \frac{1}{n} - \frac{1}{2n^2} + O(n^{-3}))\right)\cr &= \exp\left( n \ln n + \ln n + 1 + \frac{1}{2n} + O(n^{-2})\right) \cr &= n^{n+1} e \left(1 + \frac{1}{2n}+ O(n^{-2})\right)}$ so $ \dfrac{(n+1)^{n+1}}{n^n} = n e + \frac{e}{2} + O(1/n)$ Replacing $n$ by $n-1$, $ \dfrac{n^n}{(n-1)^{n-1}} = (n-1) e + \frac{e}{2} + O(1/(n-1)) = ne - \frac{e}{2} + O(1/n)$ So $ \dfrac{(n+1)^{n+1}}{n^n} - \dfrac{n^n}{(n-1)^{n-1}} = e + O(1/n)$ If you include more terms, you get $ \dfrac{(n+1)^{n+1}}{n^n} - \dfrac{n^n}{(n-1)^{n-1}} = e + \dfrac{e}{24 n^2} + \dfrac{11\; e}{640 n^4} + \ldots $

Hmm, the denominators of the coefficients seem to be the almost the same as OEIS sequence A118051 http://oeis.org/A118051 which comes from the asymptotics of the inverse harmonic numbers.

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Brothers and Knox paper can be found here.
Now here is another approach. We can write $ \frac{{(n+1)}^{n+1}}{n^n} - \frac{{n}^{n}}{{(n-1)}^{n-1}}=\left(1+\frac {1}{n}\right)^n+n\left[\left(1+\frac{1}{n}\right)^n-\left(1+\frac{1}{n-1}\right)^{n-1} \right]. $ Since $ \left(1+\frac{1}{x}\right)^x=e \bigg[1-\frac{1}{2x}+\frac{11}{24x^2}-\frac{7}{16x^3}+\frac{2447}{5760x^4}-\frac{959}{2304x^5}+\frac{238\,043}{580\,608x^6}-\frac{67223}{165\,888x^7}+\frac{559\,440\,199}{1\,393\,459\,200x^8}-\frac{123\,377\,159}{309\,657\,600x^9}+\frac{29\,128\,857\,391}{73\,574\,645\,760x^{10}}-\frac{5\,267\,725\,147}{13\,377\,208\,320x^{11}}+O\left(\frac{1}{x^{12}} \right) \bigg] $ we get $ \frac{{(n+1)}^{n+1}}{n^n} - \frac{{n}^{n}}{{(n-1)}^{n-1}}=e+\frac{e}{24n^2}+\frac{11e}{640n^4}+\frac{5525e}{580\,608n^6}+\frac{1\,212\,281e}{199\,065\,600n^8}+\frac{772\,193e}{181\,665\,792n^{10}}+O\left(\frac{1}{n^{11}}\right). $ The denominators of the coefficients $24, 640, 580\,608,199\,065\,600$ are the same as OEIS sequence, but $181\,665\,792$ is different.

What is interesting (at least for me), that the odd powers are missing and the coefficients are positive numbers.

One more thing.

If $\lim a_n=a$ and $\lim n(a_n-a_{n-1})=0$ then $\lim (n+1)a_n-na_{n-1}=a$. The special case $a_n=\left(1+\frac{1}{n}\right)^n$ is the original statement.