The number $\sum_{j=1}^{\infty} \frac{1}{j(j+1)}$ of terms closely related to 1, and find out how many ways it can be represented? (The American Mathematical Monthly) Vol. 56, No. 872

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Prove $\because \frac{1}{j(j+1)}=\frac{1}{j}-\frac{1}{j+1}$,
$$
\therefore \sum_{j=0}^{b-1} \frac{1}{j(j+1)}=\frac{1}{a}-\frac{1}{b} \text {. }
$$

Thus, the problem is equivalent to finding positive integers $a, b$, such that for a fixed integer $m>1$ we have
$$
\frac{1}{a}-\frac{1}{b}=\frac{1}{m} .
$$

From (1), the solution is clearly $a=m-1, b=m(m-1)$.
Therefore, $\frac{1}{m}=\sum_{m-1}^{m(m-1)-1} \frac{1}{j(j+1)}$.
The number of different expressions for $\frac{1}{m}$ is the number of different positive integer solutions to (3), denoted as $P(m)$. Since $\frac{1}{a}>\frac{1}{m}$, we have $a<m$. Let $a=m-c$, substituting into (3) gives $b=\frac{m^{2}}{c}-m$.

For each value of $c$, when $c \mid m^{2}$ and $1 \leqslant c<m$, there is exactly one pair of $a, b$ values that satisfy (3), and because $a<b$, they are clearly not repeated. Now consider all $d\left(m^{2}\right)-1$ factors of $m^{2}$ that are different from $m$. If $c$ is one of them, then among $c, \frac{m^{2}}{c}$, one and only one is less than $m$. Therefore, $P(m)=\frac{1}{2}\left[d\left(m^{2}\right)-1\right]$ (where $d(n)$ represents the number of distinct prime factors of $n$).