Define three sequences:

The first sequence is $$n^n: 1,\ 4,\ 27,\ 256,\ 3125,\ 46656, \ldots$$

The second sequence is that of the ratios between adjacent members of the first series, or $$\frac{(n+1)^{n+1}}{n^n}: 4,\ \frac{27}4,\ \frac{256}{27}, \ \frac{3125}{256},\ \frac{46656}{3125},\ldots.$$

The third sequence is the difference between adjacent members of the second sequence, or $$\frac{(n+2)^{n+2}}{(n+1)^{n+1}} – \frac{(n+1)^{n+1}}{n^n}: \frac{11}{4},\ \frac{295}{108},\ \frac{18839}{6912},\ \frac{2178311}{800000},\ \ldots.$$

The third sequence converges toward e, from above, and rather quickly so. Is there a proof or explanation of why this must be so?

  • $\begingroup$ A previous editor switched the sequences to display style (centered on a separate line). I felt that the entries of the sequences were then too close to each other. So I added a bit of space by inserting \ and switched to using comma as a separator. I, too, would use a semicolon as a separator when the sequences were not on separate lines. But this is a stylistic decision, so if you don't like the way it looks, just say so. Your preference should carry more weight here. You can also use the "roll back" option in edit. $\endgroup$ – Jyrki Lahtonen Aug 28 '15 at 10:28
  • 1
    $\begingroup$ Is it correct to say that "the third sequence converges rather quickly to $e$" [emphasis added]? It is true that the first element of the third sequence is roughly within 1% of $e$ and that the fifth element of the series is within roughly 0.1% of $e$. But then it takes 14 more steps and 44 more steps, respectively, to get to within 0.01% and 0.001% of $e$. Is it proper to say that this series converges "rapidly"? Or, is it an example of a sequence whose initial value happens to be "reasonably close" to the limit? (I'm aware that the Leibniz series for $\pi$ converges much more slowly...) $\endgroup$ – Mico Aug 29 '15 at 19:17
  • 2
    $\begingroup$ Though it is only described as having been "arrived at numerically by investigating the behavior of numbers that have been raised to their own power," it is worth noting this way to calculate e can be found here as #4 Power Ratio Method. (Citation: H. J. Brothers and J. A. Knox, New closed-form approximations to the logarithmic constant e, Math. Intelligencer, 20 (1998), 25-29.) $\endgroup$ – Benjamin Dickman Aug 30 '15 at 2:46
  • $\begingroup$ @Mico, is 1,000,000,000 a large number? Or $10^{100}$? Or $10^{100^{100}}$? It's subjective. You can apply some meaning to "large" or "quickly" in context, e.g. quickly relative to other convergent sequences, but the word has no absolute meaning. $\endgroup$ – Paul Draper Aug 30 '15 at 6:04
  • 1
    $\begingroup$ @PaulDraper - Are you saying, then, that the OP's claim about "rather quick convergence" (of the third sequence) lacked context? For sure, I'm considering as context the fact that the series (the Maclaurin expantion of $e$, of course) $e(T)=\sum_{k=0}^T\frac{1}{k!}$ achieves relative precision of $0.01$% and $0.001$% for $k=6$ and $8$, respectively, and that it hits the limit of double-precision accuracy (ca $1.6\times10^{-16}$) for $k=16$. $\endgroup$ – Mico Aug 30 '15 at 7:19

Let's look at the error, assuming we know the basic $e$ limit:

$$\left ( 1+\frac1{n} \right )^n = e^{n \log{\left (1+\frac1{n} \right )}} = e^{1-\frac1{2 n} + \frac1{3 n^2}+ \frac1{4 n^3}+O\left (\frac1{n^4}\right )} = e \left [1-\frac1{2 n} + \frac{11}{24 n^2}-\frac{7}{16 n^3}+O\left (\frac1{n^4}\right ) \right ]$$


$$\begin{align} \left ( 1+\frac1{n+1} \right )^{n+1} &= e \left [1-\frac1{2 (n+1)} + \frac{11}{24 (n+1)^2}-\frac{7}{16 (n+1)^3}+O\left (\frac1{n^4}\right ) \right ] \\ &= e \left [1-\frac1{2 n} + \frac{23}{24 n^2}-\frac{89}{48 n^3}+O\left (\frac1{n^4}\right ) \right ] \end{align}$$

Thus, the OP's sequence looks like, for large $n$:

$$\left(1+\frac 1{n+1}\right)^{n+1}+(n+1)\left[\left(1+\frac 1{n+1}\right)^{n+1}-\left(1+\frac 1{n}\right)^n\right] = e + \frac{e}{24 n^2} + O\left (\frac1{n^3}\right )$$

The error decreases more rapidly as one would expect - the $O(1/n)$ term vanishes. Furthermore, the sequence approaches $e$ from above rather than below, as observed.


The original version of this answer was incorrect. Amazing that nobody downvoted and the wrong answer got 10 upvotes. It should have been apparent to me that one needs to expand out to $O(1/n^3)$ to get the correct behavior. I think @robjohn saw this and got to the correct answer first, but he was too polite to mention this in my answer.

  • $\begingroup$ I apologize and corrected an error in the coefficient of the $1/n^2$ term in the answer. It is still positive of course. Answer has been checked against a log-log plot of the sequence for large n and the slope of -3 checks out. $\endgroup$ – Ron Gordon Aug 28 '15 at 21:34
  • $\begingroup$ I stumbled upon this in a recent problem recently. It's interesting, actually: $(1+1/n)^n$ is too small, but $(1+1/n)^{n+1}$ is actually too large, so that $(1+1/(n+1))^{n+1}$ is a significantly better approximation than either. Yeah, here it was: math.stackexchange.com/questions/1400932/… $\endgroup$ – Ian Aug 28 '15 at 22:13
  • $\begingroup$ I guess one way to see this intuitively is to think about $(1+x)^{1/x}$: when you change $x$ from $1/n$ to $1/(n+1)$, the resulting change is only of order $1/n^2$, so if $(1+x)^{1/x}$ has bounded derivative for small $x$, then this is somehow expected. And indeed it does; after the fact we can see it because it is $\exp(\ln(1+x)/x)$ and $\ln(1+x)/x$ turns out to have a smooth extension to $x=0$. $\endgroup$ – Ian Aug 28 '15 at 22:17
  • $\begingroup$ @Ian: $\left(1+\frac1{n+1}\right)^{n+1}$ is the same sequence as $\left(1+\frac1n\right)^n$. For faster convergence to $e$, try $\left(1+\frac1n\right)^{n+\frac12}$. It is asymptotically $e\left(1+\frac1{12n^2}\right)$ $\endgroup$ – robjohn Aug 29 '15 at 0:04
  • $\begingroup$ "more rapidly as you would expect" you probably meant "more rapidly than you would expect" (influenced by German?) $\endgroup$ – Marc van Leeuwen Aug 30 '15 at 4:25

Elementary Approach

Here is an elementary approach that uses nothing more than Bernoulli's inequality, and Bernoulli's Inequality can be proven simply with induction as shown at the end of this answer. $$ \begin{align} &\frac{(n+2)^{n+2}}{(n+1)^{n+1}}\ –\ \frac{(n+1)^{n+1}}{n^n}\\ &=(n+2)\left(1+\frac1{n+1}\right)^{n+1}-(n+1)\left(1+\frac1n\right)^n\\ &=\left(1+\frac1{n+1}\right)^{n+1}+(n+1)\left[\left(1+\frac1{n+1}\right)^{n+1}-\left(1+\frac1n\right)^n\right]\tag{1} \end{align} $$ Bernoulli's Inequality says that for $x\gt-1$, $$ (1+x)^n\ge1+nx\tag{2} $$ therefore, $$ \left(1-\frac{x}{1+x}\right)^n\ge1-n\frac{x}{1+x}\tag{3} $$ and taking reciprocals, $$ (1+x)^n\le\frac1{1-n\frac{x}{1+x}}\tag{4} $$ Setting $x=-\frac1{n^2}$ in $(2)$ and $(4)$ shows that $$ 1-\frac1n\le\left(1-\frac1{n^2}\right)^n\le1-\frac1{n+1}\tag{5} $$ Consider the quantity in square brackets from $(1)$: $$ \begin{align} \left(1+\frac1{n+1}\right)^{n+1}-\left(1+\frac1n\right)^n &=\left[\left(\frac{(n+2)n}{(n+1)^2}\right)^{n+1}-\frac{n}{n+1}\right]\left(1+\frac1n\right)^{n+1}\\ &=\left[\left(1-\frac1{(n+1)^2}\right)^{n+1}-\frac{n}{n+1}\right]\left(1+\frac1n\right)^{n+1}\tag{6} \end{align} $$ Using $(5)$, we get $$ 0\le\left(1-\frac1{(n+1)^2}\right)^{n+1}-\frac{n}{n+1}\le\frac1{(n+1)(n+2)}\tag{7} $$ In this answer, it is shown that $\left(1+\frac1n\right)^{n+1}$ is decreasing in $n$ (again using Bernoulli's Inequality). Therefore, for $n\ge1$, $$ \left(1+\frac1n\right)^{n+1}\le4\tag{8} $$ Combining $(6)$, $(7)$, and $(8)$, we get the following estimate: $$ 0\le(n+1)\left[\left(1+\frac1{n+1}\right)^{n+1}-\left(1+\frac1n\right)^n\right]\le\frac4{n+2}\tag{9} $$ Combining $(1)$ and $(9)$, we get that $$ \lim_{n\to\infty}\left(\frac{(n+2)^{n+2}}{(n+1)^{n+1}}\ –\ \frac{(n+1)^{n+1}}{n^n}\right)=e\tag{10} $$

Asymptotic Expansion

We can compute the asymptotic expansion $$ \begin{align} \frac{(n+1)^{n+1}}{n^n} &=n\left(1+\frac1n\right)^{n+1}\\ &=n\exp\left((n+1)\log\left(1+\frac1n\right)\right)\\ &=n\exp\left((n+1)\left(\frac1n-\frac1{2n^2}+\frac1{3n^3}-\frac1{4n^4}+O\left(\frac1{n^5}\right)\right)\right)\\ &=n\exp\left(1+\frac1{2n}-\frac1{6n^2}+\frac1{12n^3}+O\left(\frac1{n^4}\right)\right)\\ &=ne\left(1+\frac1{2n}-\frac1{24n^2}+\frac1{48n^3}+O\left(\frac1{n^4}\right)\right)\\ &=e\left(n+\frac12-\frac1{24n}+\frac1{48n^2}+O\left(\frac1{n^3}\right)\right)\tag{11} \end{align} $$ Therefore, $$ \frac{(n+2)^{n+2}}{(n+1)^{n+1}} =e\left(n+\frac32-\frac1{24n}+\frac1{16n^2}+O\left(\frac1{n^3}\right)\right)\tag{12} $$ Subtracting gives $$ \frac{(n+2)^{n+2}}{(n+1)^{n+1}}-\frac{(n+1)^{n+1}}{n^n} =e\left(1+\frac1{24n^2}+O\left(\frac1{n^3}\right)\right)\tag{13} $$

  • $\begingroup$ Thank you - your answer spurned me to do a numerical check and yours checked out. $\endgroup$ – Ron Gordon Aug 28 '15 at 21:35
  • $\begingroup$ How did you get $1-\frac1{n+1}$ from letting $x=-\frac1{n^2}$ in (4)? $\endgroup$ – mysatellite Sep 2 '15 at 3:48
  • $\begingroup$ @Sky: $$\begin{align}\frac1{1-n\frac{-\frac1{n^2}}{1-\frac1{n^2}}} &=\frac1{1+\frac{n}{n^2-1}}\\ &=\frac{n^2-1}{n^2+n-1}\\ &=1-\frac{n}{n^2+n-1}\\ &\le1-\frac1{n+1}\end{align}$$ $\endgroup$ – robjohn Sep 2 '15 at 6:12

Note you are computing $$(n+2)\left(1+\frac 1{n+1}\right)^{n+1}-(n+1)\left(1+\frac 1{n}\right)^n=$$$$=\left(1+\frac 1{n+1}\right)^{n+1}+(n+1)\left(\left(1+\frac 1{n+1}\right)^{n+1}-\left(1+\frac 1{n}\right)^n\right)$$

You should recognise the terms in brackets to powers $n,n+1$ as converging to $e$. The second term - the difference multiplied by $n+1$ doesn't obviously go to zero - at least not immediately. Your analysis suggests that the error term decreases more rapidly than when you have the first term alone - and that in itself is interesting.

Others will perhaps analyse the rate of convergence, or you can try that yourself.

  • 3
    $\begingroup$ But don't you need to argue why the difference in the outer parentheses goes to zero faster than $n+1$ increases? $\endgroup$ – joriki Aug 28 '15 at 10:24
  • $\begingroup$ @joriki Indeed that is so - I had intended my comment to imply that, but it isn't clear enough. I really wanted this as a comment, but the equations come out more clearly in an answer. $\endgroup$ – Mark Bennet Aug 28 '15 at 10:26

This is a continuation of @Mark Bennet's answer in which I will show, why the limit is indeed $e$.

We will concentrate on the term $T_n=(n+1)\left(\left(1+\frac{1}{n+1}\right)^{n+1}-\left(1+\frac{1}{n}\right)^n\right)$.

Firstly, we make a few modifications: $$ T_n=(n+1)\left(\left(1+\frac{1}{n+1}\right)^{n+1}-\left(1+\frac{1}{n}\right)^n\right)=\\ (n+1)\left(1+\frac{1}{n+1}\right)\left(1+\frac{1}{n+1}\right)^{n}-(n+1)\left(1+\frac{1}{n}\right)^n=\\ (n+2)\left(1+\frac{1}{n+1}\right)^{n}-(n+1)\left(1+\frac{1}{n}\right)^n=\\ \underbrace{\left(1+\frac{1}{n+1}\right)^{n}}_{u_n}+\underbrace{(n+1)\left(\left(1+\frac{1}{n+1}\right)^{n}-\left(1+\frac{1}{n}\right)^n\right)}_{v_n} $$ As we can see quite easily, $u_n$ converges to $e$. For $v_n$ we use the well known factorization $a^n-b^n=(a-b)\cdot\sum_{r=0}^{n-1}a^rb^{n-1-r}$: $$ v_n=(n+1)\left(\left(1+\frac{1}{n+1}\right)^{n}-\left(1+\frac{1}{n}\right)^n\right)=\\ (n+1)\left(\frac{1}{n+1}-\frac{1}{n}\right)\sum_{r=0}^{n-1}\left(1+\frac{1}{n+1}\right)^r\left(1+\frac{1}{n}\right)^{n-1-r}=\\ -\frac{1}{n}\sum_{r=0}^{n-1}\left(1+\frac{1}{n+1}\right)^r\left(1+\frac{1}{n}\right)^{n-1-r} $$ With the obvious inequalities $\left(1+\frac{1}{n+1}\right)^r≤\left(1+\frac{1}{n}\right)^r$ and $\left(1+\frac{1}{n}\right)^{n-1-r}≥\left(1+\frac{1}{n+1}\right)^{n-1-r}$ we obtain: $$ \left(1+\frac{1}{n+1}\right)^{n-1}=\frac{1}{n}\sum_{r=0}^{n-1}\left(1+\frac{1}{n+1}\right)^r\left(1+\frac{1}{n+1}\right)^{n-1-r}≤-v_n\\ ≤\frac{1}{n}\sum_{r=0}^{n-1}\left(1+\frac{1}{n}\right)^r\left(1+\frac{1}{n}\right)^{n-1-r}=\left(1+\frac{1}{n}\right)^{n-1} $$ Both the upper and the lower bound of $-v_n$ converge to $e$, so by the squeeze theorem, $v_n$ converges to $-e$. Therefore: $$ \lim_{n\to\infty}T_n=\lim_{n\to\infty}u_n+v_n=\lim_{n\to\infty}u_n+\lim_{n\to\infty}v_n=e-e=0 $$ This completes the answer of Mark Bennet with sufficient precision and we can conclude that your limit indeed is $e$.


As $n$ tends to $\infty$, the second sequence behaves like a line with slope $e$: $$ \frac{(n+1)^{n+1}}{n^n} = (n+1)\underbrace{\left( 1 + \frac{1}{n}\right)^n}_{\approx\ e} \approx en + e. $$ When you take the difference between adjacent points on a line, you get the slope -- in this case, $e$.


This is more like an heuristic but globally works to get hints on such defined series behavior :

Let $(U_{n})_{n\in\mathbb N}$ be your $\frac{(n+1)^{n+1}}{(n)^{n}}$ so that $(U_{n})_{n\in\mathbb N}= F(n) $ if $F : x \mapsto e ^{(x+1)\ln(x+1)-x\ln(x)}$

Then your trying to study $(U_{n+1} -U_{n})$ which behave like $F'$. Such an approach lacks rigor but permits to understand well some results on series (for instance Abel transformation). Compute $F'$ and see that its limit at infinity is $e$.

  • 4
    $\begingroup$ Couldn't you make this argument rigorous by observing that (by the Intermediate Value Theorem), $U_{n+1} - U_{n} = F'(c_n)$, where $c_n \in (n, n+1)$? $\endgroup$ – Strants Aug 29 '15 at 1:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.