给定 $n \in \mathbb N^{+}$，$n \geqslant 2026$，定义数列 $\left\{x_k\right\},\left\{y_k\right\}$，满足 $x_0=y_0=1$，且对任意 $k \in \mathbb N^{+}$，均有\[x_k=x_{k-1}^2+\frac{1}{2^n},\quad y_k=y_{k-1}^2+\frac{1}{2^n-n},\]求证：$x_n<\mathrm e < y_n$．
解法一
根据题意，$\{x_k\},\{y_k\}$ 均单调递增．设 $a=\frac{1}{2^n},b=\frac{1}{2^n-n}$，则\[\ln x_k=\ln\left(x_{k-1}^2\left(1+\frac{a}{x_{k-1}^2}\right)\right)=2\ln x_{k-1}+\ln\left(1+\frac{a}{x_{k-1}^2}\right),\]于是\[\ln x_k<2\ln x_{k-1}+\frac{a}{x_{k-1}^2}<2\ln x_{k-1}+a,\]因此\[\ln x_k<2^{k-1}(\ln x_0+a)-a<1-a<1,\]所以 $x_n<\mathrm e$． 类似可得\[\ln y_k<2^{k-1}(\ln y_0+b)-b=b\left(2^{k-1}-1\right),\]而\[ \ln y_k=2\ln y_{k-1}+\ln\left(1+\frac{b}{y_{k-1}^2}\right),\]于是\[\begin{split} \ln y_n&=\sum_{k=1}^n2^{n-k}\ln\left(1+\frac{b}{y_{k-1}^2}\right)\\ &>\sum_{k=1}^n2^{n-k}\left(\frac{b}{y_{k-1}^2}-\frac{b^2}{2y_{k-1}^4}\right)\\ &=\sum_{k=1}^n\dfrac{b\cdot 2^{n-k}}{\mathrm e^{2b(2^{k-1}-1)}}-2^n\cdot \frac{b^2}2\sum_{k=1}^n2^{-k}\\ &=2^n\cdot b\cdot \mathrm e^{2b}\sum_{k=1}^n\left(2^{-k}\cdot \mathrm e^{-b\cdot 2^k}\right)-\frac{b^2}2(2^n-1)\\ &>2^n\cdot b\cdot \mathrm e^{2b}\sum_{k=1}^n\left(2^{-k}(1-b\cdot 2^k)\right)-\frac{b^2}2\cdot 2^n\\ &=(1+\varepsilon_n)(1+2b)\left(1-\frac{1}{2^n}-\varepsilon_n\right)-\frac{b}2\varepsilon_n\\ &=1+\frac 32b\left(1-(n+2\varepsilon_n)\varepsilon_n\right)+\left(\frac{1+2b}{2^n}+\frac b2\right)(1-\varepsilon_n) \\ &>1,\end{split}\]其中 $\varepsilon_n=bn$，而 $1+\varepsilon_n=b\cdot 2^n$，因此 $y_n>\mathrm e$，命题得证．
解法二
根据题意有 $\{x_k\},\{y_k\}$ 均单调递增，记 $s=\frac{1}{2^n}$，则有\[x_k+s=x_{k-1}^2+2s\leqslant x_{k-1}^2+2s\cdot x_{k-1}<(x_{k-1}+s)^2,\]于是\[x_n+s<(x_0+s)^\frac 1s=\mathrm e,\]因此 $x_n<\mathrm e$． 对于右边，记 $s=\frac{1}{2^n-n}$，考虑构造满足关于 $k$ 递推的不等式\[y_k\geqslant 1+a_k\cdot s+b_k\cdot s^2,\]则\[y_{k+1}\geqslant (1+a_k\cdot s+b_k\cdot s^2)^2+s=1+(2a_k+1)s+(a_k^2+2b_k)s^2+2a_kb_k\cdot s^3+b_k^2\cdot s^4,\]于是定义 $a_0=b_0=1$，且\[a_{k+1}=2a_k+1,\quad b_{k+1}=a_k^2+2b_k,\]可得\[a_k=2^k-1,\quad b_k=2^kb_0+\sum_{i=0}^{k-1}2^{k-1-i}a_i^2=2^{2k+1}-(2k+1)2^k-1,\]于是\[y_k>1+\frac{2^k-1}{2^n-n}+\dfrac{2^{2k+1}-(2k+1)2^k-1}{(2^n-n)^2}\implies y_{n-1}>1+\frac 12+\frac 14=\frac {7}4,\]于是\[y_n>y_{n-1}^2=\frac{49}{16}>\mathrm e,\]综上所述，原命题得证．