One, (40 points) Find the smallest integer $c$, such that there exists a sequence of positive integers $\left\{a_{n}\right\}(n \geqslant 1)$ satisfying:
$$
a_{1}+a_{2}+\cdots+a_{n+1}<c a_{n}
$$

for all $n \geqslant 1$.
Given the problem, we have
$$
c>\frac{a_{1}+a_{2}+\cdots+a_{n+1}}{a_{n}}.
$$

For any \( n \geqslant 1 \), we have
$$
\begin{array}{l}
nc > \frac{a_{1}+a_{2}}{a_{1}} + \frac{a_{1}+a_{2}+a_{3}}{a_{2}} + \cdots + \frac{a_{1}+a_{2}+\cdots+a_{n+1}}{a_{n}} \\
= n + \frac{a_{2}}{a_{1}} + \left(\frac{a_{1}}{a_{2}} + \frac{a_{3}}{a_{2}}\right) + \cdots + \\
\left(\frac{a_{1}}{a_{n}} + \frac{a_{2}}{a_{n}} + \cdots + \frac{a_{n+1}}{a_{n}}\right) \\
= n + \left(\frac{a_{2}}{a_{1}} + \frac{a_{1}}{a_{2}}\right) + \left(\frac{a_{3}}{a_{2}} + \frac{a_{2}}{a_{3}}\right) + \cdots + \\
\left(\frac{a_{n}}{a_{n-1}} + \frac{a_{n-1}}{a_{n}}\right) + \frac{a_{n+1}}{a_{n}} + \frac{a_{1}}{a_{3}} + \\
\frac{a_{1}+a_{2}}{a_{4}} + \cdots + \frac{a_{1}+a_{2}+\cdots+a_{n-2}}{a_{n}} \\
\geqslant n + 2(n-1) + \frac{a_{n+1}}{a_{n}} + \frac{a_{1}}{a_{3}} + \\
\frac{a_{1}+a_{2}}{a_{4}} + \cdots + \frac{a_{1}+a_{2}+\cdots+a_{n-2}}{a_{n}}.
\end{array}
$$

If there exists \( n \geqslant 1 \) such that \( \frac{a_{n+1}}{a_{n}} \geqslant 2 \), then
$$
\begin{array}{l}
nc > n + 2(n-1) + 2 = 3n \\
\Rightarrow c > 3 \Rightarrow c \geqslant 4.
\end{array}
$$

If for any \( n \geqslant 1 \), we have \( \frac{a_{n+1}}{a_{n}} < 2 \), then
$$
\begin{array}{l}
nc > n + 2(n-1) + \frac{1}{2^2} + \frac{1}{2^2} + \cdots + \frac{1}{2^2}.
\end{array}
$$

Taking \( n-2 > 2^2 \), we get
$$
nc > n + 2(n-1) + 2 = 3n \\
\Rightarrow c \geqslant 4.
$$

Thus, the smallest integer \( c \) is 4.
Below is a construction: take \( a_{n} = 2^{n-1} \),
$$
\begin{array}{l}
a_{1} + a_{2} + \cdots + a_{n+1} = 1 + 2 + \cdots + 2^n \\
= 2^{n+1} - 1 \\
4a_{n} = 4 \times 2^{n-1} = 2^{n+1},
\end{array}
$$

Clearly, \( a_{1} + a_{2} + \cdots + a_{n+1} < 4a_{n} \) holds.