Where $\{x\}$ is the fractional part of $x$.

According to Desmos, with $I_n=\int\limits_{0}^{1}\{x+\frac{1}{2}\{x+\frac{1}{3}\{x+...\frac{1}{n}\}\}\}dx$

$$\begin{array}{c|c} n & I_n \\ \hline 2 & \frac{1}{2} \\ \hline 3 & \frac{19}{36} \\ \hline 4 & \frac{107}{240} \\ \hline 5 & \frac{395473}{836400} \\ \hline 6 & \approx0.459451666657 \\ \hline 7 & \approx0.445305522347 \end{array}$$

It seems to be bounded by $0.44<I_\infty<0.48$. I do not even know to how to go about proving the convergence of this because I have not found a pattern in $\frac{1}{2},\frac{19}{36},\frac{107}{240},\frac{395473}{836400}$.

Edit: I came up with this problem while messing around on Desmos. Here is the link to the graph: https://www.desmos.com/calculator/fgikxrptj2

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    $\begingroup$ If there were no $\{\;\}$ inside the first integral part, the limit of the inside expression would be $xe$, and $\int^1_0\{e x\}\,dx\approx 0.4627786...$ more or less in accordance to your bounds. I think the trick is to show that $f_n(x)=f_{n-1}(x+\frac1n\{x\})$, where $f_1(x)=\{x\}$ converges to $\frac\{ex\}$ or something like that. $\endgroup$
    – Mittens
    Feb 6 at 0:03
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    $\begingroup$ IIf the OP can also post the provenance of this problem (a compendium of problems, a statement in a paper, an old academic or contest exam) to give credit to the authors of the problem, that would be appreciated. $\endgroup$
    – Mittens
    Feb 6 at 0:13
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    $\begingroup$ @Mittens I came up with it myself. I don't suppose I was the first person to think of this but, as you can probably tell by a lot of other questions I ask on this site, I mess around a lot on desmos with the fractional part function. $\endgroup$ Feb 6 at 0:17
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    $\begingroup$ @Mittens I find Dylan's questions interesting. I don't think a lot of people need a provenance to answer his question. $\endgroup$
    – Sam
    Feb 6 at 0:21
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    $\begingroup$ @Sam: Bexacuse there are other places such as the artofproblemsolving.com where the solution might have been posted and also because is good practice to give credit. $\endgroup$
    – Mittens
    Feb 6 at 0:26

2 Answers 2


Let $f_j(x) =\left\{x + \frac{1}{2} \left\{x + \cdots + \frac{1}{j}\left\{x + \frac{1}{j+1}\right\}\cdots\right\}\right\}$ for $j > 0$. Define $$g_0 = 0; \quad g_j(x) = \left\{x + \frac{1}{2} \left\{x + \cdots + \frac{1}{j}\left\{x\right\}\cdots\right\}\right\} \text{ for } j > 0.$$ Finally, define the sequence $$ r_k = \frac{1}{(k+1)!\sum_{n=k+1}^{\infty} \frac{1}{n!}} $$ for $k \geq 0$, and write $\alpha = e - 1$ for convenience. Note $r_k$ is increasing and that $\lim_{k \to \infty} r_k = 1$.

Lemma. $f_j \to \alpha(x - r_m) + g_n(r_m)$ pointwise on $(r_m, r_{m+1})$.

Proof. Denote the domain of continuity of a function $a \colon \mathbb{R} \to \mathbb{R}$ by $$\mathcal{C}(a) = \{x \in [0,1] \mid a \text{ is continuous at } x \}.$$ Note that

  • $\mathcal{C}(a \circ b) = \mathcal{C}(b) \cap b^{-1}[\mathcal{C}(a)]$ for $a,b \colon \mathbb{R} \to \mathbb{R}$, and
  • $\mathcal{C}(a + b) = \mathcal{C}(a)$, if $b$ is continuous.

Of interest to us is the fact that $\mathcal{C}(x \mapsto \{x\}) = \mathbb{R} - \mathbb{Z}$. Now, given a function $a \colon \mathbb{R} \to \mathbb{R}$, define the $\lambda$-crease of $a$ to be the map $T_\lambda[a]$ taking $x \mapsto \frac{1}{\lambda}\{x + a(x)\}$. Then we may compute $$\mathcal{C}(T_{\lambda}[a]) = \mathcal{C}(a) \cap \{x \mid x + a(x) \not\in \mathbb{Z}\}.$$ Observe that $f_j$ is actually the result of performing multiple creases on $a \equiv \frac{1}{j+1}$. Specifically, $f_j = T_1 T_{2} \cdots T_{j} \left[\frac{1}{j+1}\right]$. Then it follows that $$\mathcal{C}(f_j) = \left\{ x \bigm| x + T_{k} \cdots T_{j}\left[\tfrac{1}{j+1}\right] < 1 \text{ for all } 1 \leq k \leq j \right\}.$$

Now, we can determine all the points of discontinuity from solutions to the equations $$f_{j, k}(x) \overset{\text{def}}{=} x + T_{k} \cdots T_{j}\left[\tfrac{1}{j+1}\right] = 1; \quad f_{j, j+1} \overset{\text{def}}{=} x + \tfrac{1}{j+1} = 1.$$ I claim that $f_{j,k}(x) < 1$ for all $x \in (0, r_{k-3})$. To see this, write $f_{j,k}(x) = x + \frac{1}{k}\{f_{j, k-1}(x)\}$, and note that for $x < r_{k-3}$, we have $$f_{j,k}(x) < r_{k-3} + \tfrac{1}{k} < 1,$$ since $$\frac{1}{r_{k-3}} = \frac{1}{k-1}\left(k + \frac{1}{k} + \frac{1}{k(k+1)} + \frac{1}{k(k+1)(k+2)}+\cdots\right) > \frac{k}{k-1}.$$

As $(r_k)_{k \in \mathbb{N}}$ is increasing, it follows that for any integer $k' \in \{k, \ldots, j\}$, one also has that $f_{j, k'}(x) < 1$ for all $x \in (0, r_{k-3})$. Then if $m(k, x)$ is the greatest integer such that $f_{j, m}(x) = 1$, we must have $m(k,x) < k \leq j$ whenever $x \in (0, r_{k-3})$. In that case, we have \begin{align*} f_{j,j+1}(x) &= x + \tfrac{1}{j+1} \\ f_{j,j}(x) &= x + \tfrac{1}{j}\{f_{j, j}(x)\} = x + \tfrac{1}{j}\left( x + \tfrac{1}{j+1} \right) \\ &{\hspace{6pt}\vdots} \\ f_{j, m}(x) &= x + \tfrac{1}{m}\left(x + \tfrac{1}{m+1}\left(x + \cdots + \tfrac{1}{j} \left(x + \tfrac{1}{j+1}\right) \cdots \right) \right) = 1. \end{align*} This is a linear equation with a unique solution $s_{j, m}$, which may or may not be in the desired interval. Therefore, there are at most $k$ points of discontinuity of $f_j$ in the interval $(0, r_{k-3})$. However, by explicitly calculating these solutions for large $j$, we can determine that exactly $k - 2$ points of discontinuity lie in this interval. In particular, we have that for $m < k$, $$s_{j,m} = \frac{\frac{1}{(m-1)!} - \frac{1}{(j+1)!}}{\frac{1}{(m-1)!} + \frac{1}{m!} + \cdots + \frac{1}{j!}} \to r_{m-2},$$ and moreover, that $s_{j,m}$ is increasing in $j$, since $\frac{c}{d} > \frac{a}{b}$ implies $\frac{a+c}{b+d} > \frac{a}{b}$ and $$\frac{\frac{1}{(j+1)!} - \frac{1}{(j+2)!}}{\frac{1}{(j+1)!}} = 1 - \frac{1}{j+2} > 1-\frac{1}{m+3} > s_{j,m}.$$ Therefore, for large $j$, the only points of discontinuity of $f_j$ in the interval $(0, r_{k-3})$ are the roots $s_{j,2}, \ldots, s_{j,k-1}$. By a similar argument, we may determine the points of discontinuity of $g_j$, and realize that they do not include any $r_{k}$. This can be done through wrangling several inequalities, but the quickest way to see this is that the points of discontinuity of $g_j$ will all be rational, whereas each $r_k$ is irrational.

We can now compute the pointwise limit of $f_j$. Suppose that $x_0 \in (r_m, r_{m+1})$ Then there exists some integer $J \geq m + 3$ such that for all $j \geq J$, we have $x_0 \in (s_{j, m+2}, s_{j, m+3})$. Since $f_j$ is right-continuous and linear on $[s_{j, m+2}, s_{j, m+3})$, we may write $$f_j(x) = \alpha_j (x - s_{j, m+2}) + f_j(s_{j, m+2}) = \alpha_j (x - s_{j, m+2}) + g_{m}(s_{j, m+2}),$$ where $\alpha_j = \sum_{n=1}^{j} \frac{1}{n!}$ is the slope of $f_j$. Since $g_m$ is continuous at $r_m$, we may take the limit to obtain

$$\lim_{j \to \infty} f_j(x) = \alpha (x - r_m) + g_m(r_m),$$ as desired.

The bounded convergence theorem implies that the limit of $I_j = \int_{0}^{1} f_j(x) \, \mathrm{d}x$ exists, and is equal to $\int_{0}^{1} f(x) \, \mathrm{d}x$. By the previous lemma, we may integrate $f$ over each interval $(r_m, r_{m+1})$, yielding $$ \lim I_j = \int_{0}^{1} f(x) \, \mathrm{d}x = \frac{1}{2\alpha} + \sum_{n=0}^{\infty} \left[g_n(r_{n})\left(r_{n+1} - r_n\right) + \frac{\alpha}{2}(r_{n+1}-r_n)^2\right]. $$


This is not a complete answer, but it will help you get started. Define $I_n = \int_0^{1} g_n(x) dx$ and $f(x) = \{x\}$ be the fractional part function. Since $f(x) = x - \lfloor x \rfloor$ (Subtracting an integer from $x$), the function $g_n$ can be simplified as $$ g_n(x) = x\left( \sum_{k=1}^{n-1} \frac{1}{k!}\right) + \frac{1}{n!} - \sum_{i=1}^{N(n)} c_i\chi_{[a_{i-1},a_i)}(x)$$ where $\{a_i\}_{i=0}^{N}$ is a partition of $[0,1]$, $0 = a_0 < a_1 < \cdots < a_N = 1$, $\chi_{[a_{i-1},a_i)} = 1$ if $x \in [a_{i-1},a_i)$ and $0$ otherwise (these are called indicator functions), and coefficients $0\le c_i < e-1$ .We also know that $a_1 = \frac{1}{\sum_{k=1}^{n-1}k!}\left(1-\frac{1}{n!}\right)$, $a_2 = \frac{1}{\sum_{k=2}^{n-1}\frac{1}{k!}}\left(\frac{1}{2}-\frac{1}{n!}\right)$ (unless $n=2$, in which case $a_2=1$) and $c_1 =0, c_2 = 1, c_3= \frac{1}{2}$. This implies that $$g_n(x) =\begin{cases} x\left( \sum_{k=1}^{n-1} \frac{1}{k!}\right) + \frac{1}{n!},& \quad 0 \le x < \frac{1}{\sum_{k=1}^{n-1}k!}\left(1-\frac{1}{n!}\right) \\ x\left(\sum_{k=1}^{n-1}\frac{1}{k!}\right)+\frac{1}{n!}-1,& \frac{1}{\sum_{k=1}^{n-1}k!}\left(1-\frac{1}{n!}\right)\le x<\frac{1}{\sum_{k=2}^{n-1}\frac{1}{k!}}\left(\frac{1}{2}-\frac{1}{n!}\right)\end{cases}$$ and $I_n$ can be simplified as: $$ I_n =\frac{1}{2}\left( \sum_{k=1}^{n-1} \frac{1}{k!}\right) + \frac{1}{n!} - \sum_{i=2}^{N(n)} c_i|a_i-a_{i-1}| $$ If something can be said about the sequence $c_i$ then the limiting value can be determined. In the limiting case $$I_{\infty} = \frac{e-1}{2}-\left(\frac{1}{2}\frac{1}{e-2}-\frac{1}{e-1}\right)-\sum_{i=3}^{\infty} c_i|a_i-a_{i-1}|. $$


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