Question 139, Given $p, q \in R^{+}$, if the infinite sequence $\left\{a_{n}\right\}$ satisfies: $a_{1}=1$, and for any natural numbers $m$ and $n$, $\left|a_{m+n}-a_{m}-a_{n}\right| \leq \frac{1}{p m+q n}$, find the general term $a_{n}$ of the sequence.
Question 139, Solution: Let $m=1$, then by the condition we have $\left|\left(a_{n+1}-a_{n}\right)-a_{1}\right| \leq \frac{1}{p+q n}$, thus $\lim _{n \rightarrow+\infty}\left(a_{n+1}-a_{n}\right)=a_{1}$.
Let $m=2$, then by the condition we have $\left|\left(a_{n+2}-a_{n}\right)-a_{2}\right| \leq \frac{1}{2 p+q n}$, thus $\lim _{n \rightarrow+\infty}\left(a_{n+2}-a_{n}\right)=a_{2}$.
Similarly, in general, we have $\lim _{n \rightarrow+\infty}\left(a_{n+m}-a_{n}\right)=a_{m}$.
Therefore, for any $m \in N^{+}$, we have:
$$
\begin{array}{l}
a_{m}=\lim _{n \rightarrow+\infty}\left(a_{n+m}-a_{n}\right) \\
=\lim _{n \rightarrow+\infty}\left[\left(a_{n+m}-a_{n+m-1}\right)+\left(a_{n+m-1}-a_{n+m-2}\right)+\ldots+\left(a_{n+1}-a_{n}\right)\right] \\
=\lim _{n \rightarrow+\infty}\left(a_{n+m}-a_{n+m-1}\right)+\lim _{n \rightarrow+\infty}\left(a_{n+m-1}-a_{n+m-2}\right)+\ldots+\lim _{n \rightarrow+\infty}\left(a_{n+1}-a_{n}\right) \\
=m a_{1}=m
\end{array}
$$

Therefore, the general term of the sequence $\mathrm{a}_{\mathrm{n}}=\mathrm{n}$.