[Proof] By contradiction. Suppose the conclusion to be proved does not hold, then there exists $k \in \mathbb{N}$, such that when $n \geqslant k$ we have
$$\frac{a_{1}+a_{n+1}}{a_{n}} \leqslant 1+\frac{1}{n} \text {, }$$

which implies $\frac{a_{n}}{n} \geqslant \frac{a_{1}}{n+1}+\frac{a_{n+1}}{n+1}$.
Thus, we can obtain
$$\begin{array}{c} \frac{a_{k}}{k} \geqslant \frac{a_{1}}{k+1}+\frac{a_{k+1}}{k+1} \geqslant \frac{a_{1}}{k+1}+\frac{a_{1}}{k+2}+\frac{a_{k+2}}{k+2} \\ \geqslant \cdots \geqslant a_{1} \sum_{i=1}^{m} \frac{1}{k+i}+\frac{a_{k+m}}{k+m} \geqslant \cdots \end{array}$$

Since $\lim _{m \rightarrow \infty} \sum_{i=1}^{m} \frac{1}{k+i}=+\infty$, this leads to a contradiction! Therefore, there must be infinitely many $n \in \mathbb{N}$, such that
$$\frac{a_{1}+a_{n+1}}{a_{n}}>1+\frac{1}{n} \text {. }$$