# A possible surprise involving Euler's constant $e$ [closed]

Let

\begin{align*} c_n &= n!\left(e-\sum_{k=0}^n \frac{1}{k!}\right) \\ \\ u_n &= \bigg\lfloor{\frac{1}{c_n} \bigg\rfloor} \\ \\ v_n &= \bigg\lfloor{\frac{1}{1/c_n-\lfloor{u_n} \rfloor}} \bigg\rfloor \end{align*}

Are $$u_n = n$$ and $$v_n = n+1$$ for all $$n \geq 0$$?

The question extends as follows (January 16). Let

\begin{align*} r_{1,n} &= 1/c_n\\ r_{2,n} &= 1/r_{1,n}\\ r_{3,n} &= 1/r_{2,n}\\ &\vdots\\ s_{1,n} &= \lfloor{1/\{r_{1,n}\}}\rfloor, \text{ where }\{ \} \text{ denotes fractional part}\\ s_{2,n} &= \lfloor{r_{2,n}} \rfloor \\ s_{3,n} &= \lfloor{r_{3,n}} \rfloor \\ &\vdots\\ \end{align*}

Are $$(s_{2,n})=(1,1,2,2,3,3, \ldots)$$ and $$(s_{3,n})=(2,1,3,1,4,1,6,1,7,1, \ldots)$$, where the only numbers missing are $$5+4h$$ for $$h \geq0$$? In general, does the regularity of the continued fraction for $$e$$ imply some sort of regularity for $$(s_{k,n})$$ for $$k \geq 5$$?

Informally speaking, things look very irregular for higher $$k$$, and I wonder if the method in Answer 1 extends to these deeper cases.

## closed as off-topic by abx, Wojowu, Ben McKay, Steven Landsburg, Peter LeFanu LumsdaineJan 12 at 17:21

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• Isn't this clear from the expansion $e = \sum_{k = 0}^\infty \frac{1}{k!}$? – user44191 Jan 12 at 15:59

## 1 Answer

Sure. $$c_n = \frac{1}{n+1} + \frac{1}{(n+1)(n+2)}+ \frac{1}{(n+1)(n+2)(n+3)} \cdots$$ so $$\frac{1}{n+1} < c_n < \frac{1}{n+1}+\frac{1}{(n+1)^2}+\frac{1}{(n+1)^3}+\cdots = \frac{1}{n}.$$ That proves the claim about $$\lfloor 1/c_n \rfloor$$.

To get the second equality, we need to tighten our bounds. We have $$\frac{1}{n+1} + \frac{1}{(n+1)(n+2)} + \frac{1}{(n+1)(n+2)(n+3)} + \frac{1}{(n+1)(n+2)(n+3)(n+4)} = \frac{41 + 34 n + 10 n^2 + n^3}{(n+1)(n+2)(n+3)(n+4)} < c_n\ \mbox{and}$$ $$c_n < \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} + \frac{1}{(n+1)(n+2)^2} + \cdots = \frac{n+2}{(n+1)^2}.$$ Taking reciprocals, $$\frac{(n+1)(n+2)(n+3)(n+4)}{41 + 34 n + 10 n^2 + n^3}-n =\frac{n^2+9 n+24}{n^3+10 n^2+34 n+41}> c_n^{-1}-n\ \mbox{and}$$ $$c_n^{-1}-n > \frac{(n+1)^2}{n+2} - n = \frac{1}{n+2}.$$ Doing it again, $$\frac{n^3+10 n^2+34 n+41}{n^2+9 n+24}=n+1+\frac{n+17}{n^2+9 n+24}<(c_n^{-1}-n)^{-1} < n+2$$ so $$n+1 < (c_n^{-1}-n)^{-1} < n+2.$$

One should be able to get any number of terms of the partial fraction of $$c_n$$ in this manner, limited only by one's patience.