I need help with the next problem:
Let d and $a_1$ be numbers. Define $a_j$ for each integer $j \gt 1$ by the equation $a_{j+1} = a_j + d$, and suppose that $a_j \neq 0$ for all $j \in \mathbb N$. Find the sum $$\sum_{j=1}^n \frac{1}{a_j a_{j+1}}$$ for all $n \in \mathbb N.$
My solution:
Let $a_j = a_1 + (j-1) d$ and $a_{j+1} = a1+jd$
Then
$ a_j a_{j+1} = [a_1 + (j-1) d] ·[a1+jd] = (a_1+jd-d)(a_1+jd)$
Thus
$$\sum_{j=1}^n \frac{1}{a_j a_{j+1}} = \sum_{j=1}^n \left( \frac{A}{(a_1 + jd -d)} + \frac{B}{a_1+jd}\right )$$
Solving A and B:
$$A = \frac{d-a_1}{d²}= \frac{1}{d} - B \\ B = \frac{a_1}{d²}$$
Thus
$$\sum_{j=1}^n \frac{1}{a_j a_{j+1}} = \sum_{j=1}^n \left( \frac{1}{d(a_1 + jd -d)} + \frac{B}{a_1+jd} - \frac{B}{(a_1 + jd -d)} \right ) = \sum_{j=1}^n \left( \frac{1}{d(a_1 + jd -d)} \right) + \sum_{j=1}^n \left(\frac{B}{a_1+jd} - \frac{B}{(a_1 + jd -d)} \right ) $$
Applying Telescoping Series:
$$\sum_{j=1}^n \left(\frac{B}{a_1+jd} - \frac{B}{(a_1 + jd -d)} \right ) = \frac{B}{a_1+nd} - \frac{B}{a_1}$$ But I don't know how to solve: $$\sum_{j=1}^n \frac{1}{d(a_1 + jd -d)} = \frac{1}{d}\sum_{j=1}^n \frac{1}{d(a_1 + jd -d)} = \frac{1}{d}\sum_{j=1}^n \frac{1}{a_1 + (j-1)d} = \frac{1}{d}\sum_{j=1}^n \frac{1}{a_j} $$
Thank you so much!
