TITLE: Does recursively replacing $\frac1n$ by $\frac1n(\frac12+\dots+\frac1{n+1})$ really converge to $\frac1e$?
QUESTION [22 upvotes]: I was thinking of a problem and have an answer through computer programming, but am unable to prove it mathematically. Start with the following: 
$$\frac{1}{2}\bigg(\frac{1}{2} + \frac{1}{3}\bigg)\approx 0.41666$$ 
Replace every unit fraction being summed in the parentheses (in this case, the inner 1/2 and 1/3) with 
$$\frac{1}{n}\bigg(\frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n+1}\bigg)$$
where $n$ is the denominator of the unit fraction. If we apply this process once, we get
$$\frac{1}{2}\bigg(\frac{1}{2}\bigg(\frac{1}{2} + \frac{1}{3}\bigg)+\frac{1}{3}\bigg(\frac{1}{2} + \frac{1}{3}+\frac{1}{4}\bigg)\bigg)\approx 0.3888$$
and if we apply it again, we get
$$\frac{1}{2}\bigg(\frac{1}{2}\bigg(\frac{1}{2}\bigg(\frac{1}{2} + \frac{1}{3}\bigg) + \frac{1}{3}\bigg(\frac{1}{2} + \frac{1}{3}+ \frac{1}{4}\bigg)\bigg)+\frac{1}{3}\bigg(\frac{1}{2}\bigg(\frac{1}{2} + \frac{1}{3}\bigg) + \frac{1}{3}\bigg(\frac{1}{2} + \frac{1}{3}+ \frac{1}{4}\bigg)+\frac{1}{4}\bigg(\frac{1}{2} + \frac{1}{3}+ \frac{1}{4}+ \frac{1}{5}\bigg)\bigg)\bigg)\approx 0.37754$$
As we repeat this process, the result seems to approach $e^{-1}$.
If anyone could prove this or direct me to a proof of this, that would be wonderful!

REPLY [3 votes]: We provide a recurrence relation for the numbers converging to $\frac{1}{e}$. We start with some introductory aspects.
Representation via sums
We denote the elements of the sequence which are iteratively generated by OPs rule with $\alpha_n, n\geq 2$ and derive for small $n$
\begin{align*}
\alpha_2&=\frac{1}{2}\\
\alpha_3&=\frac{1}{2}\sum_{j_1=2}^3\frac{1}{j_1}\tag{i}\\
&=\frac{1}{2}\left(\frac{1}{2}+\frac{1}{3}\right)=\frac{5}{12}=0,41\dot{6}\\
\alpha_4&=\frac{1}{2}\sum_{j_1=2}^3\frac{1}{j_1}\sum_{j_2}^{j_1+1}\frac{1}{j_2}\tag{ii}\\
&=\frac{1}{2}\left(\frac{1}{2}\sum_{j_2=2}^3\frac{1}{j_2}+\frac{1}{3}\sum_{j_2=2}^4\frac{1}{j_2}\right)\\
&=\frac{1}{2}\left(\frac{1}{2}\left(\frac{1}{2}+\frac{1}{3}\right)+\frac{1}{3}\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)\right)
=\frac{7}{18}=0,3\dot{8}\\
\alpha_5&=\frac{1}{2}\sum_{j_1=2}^3\frac{1}{j_1}\sum_{j_2}^{j_1+1}\frac{1}{j_2}\sum_{j_3=2}^{j_2+1}\frac{1}{j_3}\tag{iii}\\
&=\frac{1}{2}\left(\frac{1}{2}\sum_{j_2=2}^3\frac{1}{j_2}\sum_{j_3=2}^{j_2+1}\frac{1}{j_3}+\frac{1}{3}\sum_{j_2=2}^4\frac{1}{j_2}\sum_{j_3=2}^{j_2+1}\frac{1}{j_3}\right)\\
&=\frac{1}{2}\left(\frac{1}{2}\left(\frac{1}{2}\sum_{j_3=2}^3\frac{1}{j_3}+\frac{1}{3}\sum_{j_3=2}^4\frac{1}{j_3}\right)\right.\\
&\qquad \left.+\frac{1}{2}\left(\frac{1}{2}\sum_{j_3=2}^3\frac{1}{j_3}+\frac{1}{3}\sum_{j_3=2}^4\frac{1}{j_3}+\frac{1}{4}\sum_{j_3=2}^5\frac{1}{j_3}\right)\right)\\
&=\frac{1}{2}\left(\frac{1}{2}\left(\frac{1}{2}\left(\frac{1}{2}+\frac{1}{3}\right)+\frac{1}{3}\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)\right)\right.\\
&\qquad\left.+\frac{1}{3}\left(\frac{1}{2}\left(\frac{1}{2}+\frac{1}{3}\right)+\frac{1}{3}\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)
+\frac{1}{4}\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)\right)\right)\\
&=\frac{1\,631}{4\,320}=0,377\,54\overline{629}
\end{align*}
in accordance with OP's calculation.

From (i),(ii),(iii) we  obtain for general $n$:
  \begin{align*}
\color{blue}{\alpha_n=\frac{1}{2}\sum_{j_1=2}^3\frac{1}{j_1}\sum_{j_2=2}^{j_1+1}\frac{1}{j_2}\sum_{j_3=2}^{j_2+1}\frac{1}{j_3}
\cdots\sum_{j_{n-2}=2}^{j_{n-3}+1}\frac{1}{j_{n-2}}\sum_{j_{n-1}=2}^{j_{n-2}+1}\frac{1}{j_{n-1}}\qquad\qquad n\geq 3}\tag{1}
\end{align*}

Note: The ubiquitous Catalan numbers occur here also. If we write $\alpha_n$ as
\begin{align*}
\alpha_2&=\frac{1}{2},&C_1=1\\
\alpha_3&=\frac{1}{\color{blue}{2}\cdot 2}+\frac{1}{\color{blue}{2}\cdot 3},&C_2=2\tag{2}\\
\alpha_4&=\frac{1}{\color{blue}{2\cdot 2}\cdot 2}+\frac{1}{\color{blue}{2\cdot 2}\cdot 3}+\frac{1}{\color{blue}{2\cdot 3}\cdot 2}+\frac{1}{\color{blue}{2\cdot 3}\cdot 3}+\frac{1}{\color{blue}{2\cdot 3}\cdot 4},&C_3=5\\
&\ \,\vdots
\end{align*}
we see the number of summands of $\alpha_n$ is $C_{n-1}$. The blue marked factors in $\alpha_n$ come from the factors of $\alpha_{n-1}$.
Recurrence relation
In order to find a recurrence relation we would like to reverse the order of summation in (1). Alas this is somewhat cumbersome since the upper limit of an inner sum does not nicely correspond with the lower limit of the next outer sum. We will instead show the relationship with the help of number triangles which might also provide additional insight
$$
\begin{array}{r|cccc}
n&w_{\alpha_2}&w_{\alpha_3}&w_{\alpha_4}&w_{\alpha_5}\\
\hline\\
5&&&&\frac{1}{5}\\\\
4&&&\frac{1}{4}&\frac{1}{4}\\\\
3&&\frac{1}{3}&\frac{1}{3}&\frac{1}{3}\\\\
2&\frac{1}{2}&\frac{1}{2}&\frac{1}{2}&\frac{1}{2}\\
\\
\hline\\
\\
\end{array}
\qquad\qquad\qquad
\begin{array}{r|rrrr}
n&\alpha_2&\alpha_3&\alpha_4&\alpha_5\\
\hline\\
5&&&&\frac{1}{120}\\\\
4&&&\frac{1}{4}&\frac{12}{288}\\\\
3&&\frac{1}{6}&\frac{5}{36}&\frac{7}{54}\\\\
2&\frac{1}{2}&\frac{1}{4}&\frac{5}{24}&\frac{7}{36}\\
\\
\hline\\
\alpha_n&\frac{1}{2}&\frac{5}{12}&\frac{7}{18}&\frac{1\,631}{4\,320}
\end{array}
$$
The left table shows a triangular region of weights $\frac{1}{2},\frac{1}{3},\ldots$ with weight $\frac{1}{n}$ in row $n$. These weights are used to build the entries in the right table and $\alpha_n$ which is the sum of the entries in the $n$-th column. We denote rows and columns starting from $2$ so that the bottom left-most element is $a_{22}=\frac{1}{2}$ while the top right-most element is $a_{55}=\frac{1}{5!}=\frac{1}{120}$. We have
\begin{align*}
\color{blue}{\alpha_2}&=a_{22}\color{blue}{=\frac{1}{2}}\\
\color{blue}{\alpha_3}&=a_{32}+a_{33}\\
&=\frac{1}{2}a_{22}+\frac{1}{3}a_{22}\\
&=\frac{1}{4}+\frac{1}{6}\color{blue}{=\frac{5}{12}}\\
\color{blue}{\alpha_4}&=a_{42}+a_{43}+a_{44}\\
&=\frac{1}{2}\left(a_{32}+a_{33}\right)+\frac{1}{3}\left(a_{32}+a_{33}\right)+\frac{1}{4}a_{33}\\
&=\frac{5}{24}+\frac{5}{36}+\frac{1}{24}\color{blue}{=\frac{7}{18}}\\
\color{blue}{\alpha_5}&=a_{52}+a_{53}+a_{54}+a_{55}\\
&=\frac{1}{2}\left(a_{42}+a_{43}+a_{44}\right)+\frac{1}{3}\left(a_{42}+a_{43}+a_{44}\right)
+\frac{1}{4}\left(a_{43}+a_{44}\right)+\frac{1}{5}a_{55}\\
&=\frac{7}{36}+\frac{7}{54}+\frac{13}{288}+\frac{1}{120}\color{blue}{=\frac{1\,631}{4\,320}}
\end{align*}

From the calculations above we derive the rule to build the entries $a_{k,l}$:

Take $\frac{1}{k}$ times the sum of all entries of column $k-1$ and row greater or equal $\max\{l-1,2\}$.
The recurrence relation is
  \begin{align*}
a_{n,j}&=\frac{1}{j}\sum_{k=\max\{j-1,2\}}^{n-1}a_{n-1,k}\qquad\qquad\qquad\qquad\qquad n\geq 3,j=2,\ldots,n\\
\color{blue}{\alpha_2}&\color{blue}{=\frac{1}{2}}\\
\color{blue}{\alpha_n}&=\sum_{j=2}^na_{n,j}\color{blue}{=\sum_{j=2}^n\frac{1}{j}\sum_{k=\max\{j-1,2\}}^{n-1}a_{n-1,k}\qquad\qquad n\geq 3,j=2,\ldots,n}\tag{3}
\end{align*}


We can show the validity of (3) by successively applying the recurrence relation to obtain
\begin{align*}
\color{blue}{\alpha_n}&=\sum_{j_1=2}^na_{n,j_1}\\
&=\sum_{j_1=2}^n\frac{1}{j_1}\sum_{j_2=\max\{j_1-1,2\}}^{n-1}a_{n-1,j_2}\\
&=\sum_{j_1=2}^n\frac{1}{j_1}\sum_{j_2=\max\{j_1-1,2\}}^{n-1}\frac{1}{j_2}\sum_{j_3=\max\{j_2-1,2\}}^{n-2}a_{n-2,j_3}\\
&\ \,\vdots\\
&=\sum_{j_1=2}^n\frac{1}{j_1}\sum_{j_2=\max\{j_1-1,2\}}^{n-1}\frac{1}{j_2}
\cdots\sum_{j_{n-2}=max\{j_{n-3}-1,2\}}^{3}\frac{1}{j_{n-2}}\sum_{j_{n-1}=\max\{j_{n-2}-1,2\}}^{2}a_{2,j_{n-1}}\\
&\,\,\color{blue}{=\frac{1}{2}\sum_{j_1=2}^n\frac{1}{j_1}\sum_{j_2=\max\{j_1-1,2\}}^{n-1}\frac{1}{j_2}
\cdots\sum_{j_{n-2}=max\{j_{n-3}-1,2\}}^{3}\frac{1}{j_{n-2}}}\tag{3}
\end{align*}
and (3) is the same as (1) with the summation in reverse order.

Notes:

It is interesting, that summing up the columns in the  right table above gives a monotonically increasing sequence converging to $\frac{1}{e}$. 
  \begin{align*}
\lim_{n\to\infty} \alpha_n=\frac{1}{e}
\end{align*}
  What I find amazing is the usage of the weights $\frac{1}{j}$ used in the expression of $\alpha_n$ (see e.g. (2)) which makes them similar looking to
  \begin{align*}
\sum_{j=0}^n(-1)^j\frac{1}{j!}
\end{align*}
  but without the alternating sign.
It would be great if someone could show the convergence of $(\alpha_n)$ by solving the recurrence relation (3).
It might be interesting that the same structure of the recurrence relation (3) but with different weights, namely $n-1$ instead of $\frac{1}{n}$ and different boundary values $a_{2,n}=1$ instead of $\frac{1}{2}$ and $a_{n,n}=n-1$ instead of $\frac{1}{n}$ is archived as A185423 in OEIS denoting the number of Ordered forests of $k$ increasing plane unary-binary trees with $n$ nodes.
  $$
\begin{array}{r|cccc}
n&&&&\\
\hline\\
5&&&&3\\\\
4&&&3&3\\\\
3&&2&2&2\\\\
2&1&1&1&1\\
\end{array}
\qquad\qquad\qquad
\begin{array}{r|rrrr}
A185423&&&&\\
\hline\\
5&&&&24\\\\
4&&&6&36\\\\
3&&2&6&30\\\\
2&1&1&3&9\\
\end{array}
$$