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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.

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But it seems to me that you are using Stolz Cesaro in the opposite way. I mean as you can find on Wiki en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem, if you know that $\lim_{n\to +\infty}\frac{f(n+1)-f(n)}{n+1-n}=l$ exists, and in your case it is the statement of the theorem so you don't have the premise, then you could conclude that $\lim_{n\to+\infty}\frac{f(n)}{n}$ also exists and it is equal to $l$. But, i repeat, you are reversing the theorem, which is not an if and only if. Am I misinterpretating? –  uforoboa Sep 10 '12 at 19:24
    
@uforoboa: when $\lim_{n\to\infty}\frac{b_{n}}{b_{n+1}}$ is $1$ or doesn't exist then we cannot be sure if the converse theorem works. But in this case we knew the result from the beginning. So, it could be true or not. But keep in mind that the converse theorem may be successfully applied in many cases if that limit is a real positive number different from $1$. –  Chris's sis Sep 10 '12 at 19:39
    
@uforoboa: in major part of the cases (I admit that there may be an exception or two) where I applied the converse theorem I got the correct result even if that limit was $1$. In other words, it's about hundreds of limits where it perfectly works. Empirically speaking I had success with it, but I understand your concern and agree with what you said. –  Chris's sis Sep 10 '12 at 19:47
    
@uforoboa: I'll try to find out more information about the converse theorem when $ \lim_{n\to\infty}\frac{b_{n}}{b_{n+1}}=1$. I may suppose that some research exists on this subject. In the books I read so far I only found that in this case one cannot rely on the result. This is somehow sad since the converse theorem could save a lot of work for many limits and generally offers very fast and nice solutions. –  Chris's sis Sep 10 '12 at 20:00
    
At any rate you should not say that yours is a proof... It is misleading for unexperienced users. –  uforoboa Sep 11 '12 at 0:00

2 Answers 2

up vote 6 down vote accepted

$$\eqalign{(n+1)^{n+1} = \exp\left((n+1) \ln(n+1)\right) &= \exp\left(((n+1) (\ln n + \frac{1}{n} - \frac{1}{2n^2} + O(n^{-3}))\right)\cr &= \exp\left( n \ln n + \ln n + 1 + \frac{1}{2n} + O(n^{-2})\right) \cr &= n^{n+1} e \left(1 + \frac{1}{2n}+ O(n^{-2})\right)}$$ 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.

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