621 Find all real numbers $\lambda$ such that there does not exist an infinite positive sequence $\left\{a_{n}\right\}$ satisfying
$$
a_{n}+1 \leqslant \sqrt[n]{\lambda} a_{n-1}(n \geqslant 2)
$$

always holds.
The set of all $\lambda$ that satisfy the problem is $(0, \mathrm{e}]$.
Let $s=\ln \lambda$. Then $\lambda=\mathrm{e}$.
We will solve the problem in two steps.
First, we provide an example to show that when $s>1$, there exists an infinite positive sequence $\left\{a_{n}\right\}$ that satisfies equation (1) always holds.
Consider the sequence $\left\{b_{n}\right\}$:
$b_{1}=\frac{1}{s-1}, b_{n}-b_{n-1}=\frac{1}{(n+1)^{s}}$.
Then $b_{n}=\frac{1}{s-1}-\left(\frac{1}{3^{s}}+\frac{1}{4^{s}}+\cdots+\frac{1}{(n+1)^{s}}\right)$.
By $\sum_{k=2}^{n+1} \frac{1}{k^{s}}0$.
Let $a_{n}=(n+1)^{s} b_{n}$.
Then $a_{n}>0$, and
$\frac{a_{n}}{(n+1)^{s}}+\frac{1}{(n+1)^{s}}=\frac{a_{n-1}}{n^{s}}$,

i.e., $a_{n}+1=\left(1+\frac{1}{n}\right)^{s} a_{n-1} \leqslant\left(\mathrm{e}^{\frac{1}{n}}\right)^{s} a_{n+1}$.
Thus, $a_{n}+1 \leq \sqrt[n]{\lambda} a_{n-1}$.
Therefore, the infinite positive sequence $\left\{a_{n}\right\}$ satisfies the problem.
Second, we prove that when $s \leqslant 1$, there does not exist an infinite positive sequence $\left\{a_{n}\right\}$ that satisfies the problem.
It suffices to prove the case when $s=1$.
We will use proof by contradiction.
Assume that when $s=1$, i.e., $\lambda=\mathrm{e}$, there exists an infinite positive sequence $\left\{a_{n}\right\}$ that satisfies equation (1) always holds.
Consider the sequence $\left\{c_{n}\right\}$:
$c_{n}=\frac{a_{n-1}}{1+a_{n}} c_{n-1}\left(n \geqslant 2, c_{1}>0\right)$.
Then $c_{n}>0$.
Also, $\frac{c_{n-1}}{c_{n}}=\frac{1+a_{n}}{a_{n-1}} \leqslant \sqrt[n]{\mathrm{e}}$, i.e.,
$c_{n-1} \mathrm{e}^{-\frac{1}{n}} \leqslant c_{n}$.
Thus, $c_{n} \geqslant \mathrm{e}^{-\frac{1}{n}} c_{n-1} \geqslant \mathrm{e}^{-\frac{1}{n}-\frac{1}{n-1}} c_{n-2} \geqslant \cdots$
$\geqslant \mathrm{e}^{-\left(\frac{1}{n}+\frac{1}{n-1}+\cdots+\frac{1}{2}\right)} c_{1} \geqslant \mathrm{e}^{-\ln n} c_{1}$.
Notice that, $\sum_{i=2}^{n} \frac{1}{i} \leqslant \int_{1}^{n} \frac{1}{x} \mathrm{~d} x=\ln n$.
Thus, $c_{n} \geqslant \frac{1}{n} c_{1}$.
Therefore, $\sum_{i=1}^{n} c_{i} \geqslant c_{1}\left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right)$.
By the divergence of the harmonic series $1+\frac{1}{2}+\cdots+\frac{1}{n}+\cdots$, we know that $\sum_{i=1}^{n} c_{i}$ is unbounded.
From equation (2), we have
$c_{n}\left(1+a_{n}\right)=a_{n-1} c_{n-1}$.
Then $c_{n}=a_{n-1} c_{n-1}-c_{n} a_{n}$.
Summing up, we get $\sum_{i=2}^{n} c_{i}=a_{1} c_{1}-c_{n} a_{n}<a_{1} c_{1}$.
Thus, $\sum_{i=1}^{n} c_{i}<\left(a_{1}+1\right) c_{1}$, which contradicts the unboundedness of $\sum_{i=1}^{n} c_{i}$.
Therefore, the assumption is false.
Hence, when $s=1$, there does not exist an infinite positive sequence $\left\{a_{n}\right\}$ that satisfies equation (1) always holds.

In conclusion, when $\lambda \in(0, \mathrm{e}]$, there does not exist an infinite positive sequence $\left\{a_{n}\right\}$ that satisfies equation (1) always holds.