A way to calculate e?

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?

Solution 1:

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

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

Then

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

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.

ADDENDUM

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.

Solution 2:

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 $$ \frac{(n+2)^{n+2}}{(n+1)^{n+1}}-\frac{(n+1)^{n+1}}{n^n} =e\left(1+\frac1{24n^2}+O\left(\frac1{n^3}\right)\right)\tag{13} $$