Question 140, Find the smallest real number $\mathrm{r}$, such that there exists a sequence of positive real numbers $\left\{\mathrm{x}_{\mathrm{n}}\right\}$, satisfying for any $\mathrm{n} \in \mathrm{N}^{+}$, we have $\sum_{\mathrm{i}=1}^{\mathrm{n}+1} \mathrm{x}_{\mathrm{i}} \leq \mathrm{r} \cdot \mathrm{x}_{\mathrm{n}}$.
Question 140, Solution: Clearly, $r \in R^{+}$.
Let $S_{n}=\sum_{i=1}^{n} x_{1}$, then $\left\{S_{n}\right\}$ is a strictly increasing sequence, and $S_{n+1} \leq r \cdot\left(S_{n}-S_{n-1}\right)$. Therefore, we know that $r \cdot S_{n} \geq S_{n+1}+r \cdot S_{n-1} \geq 2 \sqrt{S_{n+1} \cdot r \cdot S_{n-1}}$, rearranging gives $S_{n+1} \leq \frac{r}{S_{n}} \leq \frac{S_{n}}{S_{n-1}}$, thus $\frac{S_{n+1}}{S_{n}} \leq\left(\frac{r}{4}\right)^{n-1} \cdot \frac{S_{2}}{S_{1}}$. If $r<4$, then as $n \rightarrow+\infty$, $\left(\frac{r}{4}\right)^{n-1} \cdot \frac{s_{2}}{s_{1}} \rightarrow 0$, so for sufficiently large $n$, there must be $\frac{s_{n+1}}{s_{n}}<1$, which contradicts the fact that $\left\{S_{n}\right\}$ is a strictly increasing sequence. Hence, $r \geq 4$.
On the other hand, when $r=4$, taking $x_{1}=2^{i-1}$, we have $\sum_{i=1}^{n+1} x_{1}=4 x_{n}$.
In conclusion, the minimum value of $r$ that satisfies the condition is 4.