# Calculate the sum of the series $\sum_{n=0}^{\infty} \frac{1}{a_{n}\cdot a_{n+1}\cdot\ldots\cdot a_{n+7}}$, where $a_k = ak + b$

Let $a_k = ak + b$; define the following series:

$$\sum_{n=0}^{\infty} \frac{1}{a_{n}\cdot a_{n+1}\cdot\ldots\cdot a_{n+7}}.$$

I have to prove that this series converges and I have to find its sum. Another question that arises is: in the statement of the problem it is not specified wheater $a,b \in \mathbb{N}$, $a,b \in \mathbb{Z}$, $a,b \in \mathbb{Q}$, or $a,b \in \mathbb{R}$, which one should I assume? Why?

I have no idea at all about what to do with this exercise. All I can do is to guess that this series will turn out to be telescoping or geometric (since I've to calculate the sum). Can you show me in detail what I should do?

• Hint: If: $\frac{-b}{a}\notin \mathbb N$, then: $$\sum_{n=0}^\infty \frac{1}{a_na_{n+1}a_{n+7}}\le \sum_{n=1}^\infty \frac{1}{n^3}$$ – hamid kamali May 20 '15 at 11:34

By using partial fraction decomposition we have: $$\frac{1}{x(x+1)\cdot\ldots\cdot(x+7)}=\frac{1}{7!}\sum_{j=0}^{7}\frac{(-1)^j \binom{7}{j}}{x+j}\tag{1}$$ hence by replacing $x$ with $\frac{b}{a}+z$ we get: $$\frac{1}{\left(z+\frac{b}{a}\right)\left(z+\frac{b+a}{a}\right)\cdot\ldots\cdot\left(z+\frac{b+7a}{a}\right)}=\frac{1}{7!}\sum_{j=0}^{7}\frac{(-1)^j \binom{7}{j}}{\frac{b}{a}+z+j}\tag{2}$$ and: $$\frac{1}{a_n\cdot a_{n+1}\cdot\ldots\cdot a_{n+7}}=\frac{1}{a^7\cdot7!}\sum_{j=0}^{7}\frac{(-1)^j \binom{7}{j}}{b+an+aj}\tag{3}$$ hence by summing both terms over $n\geq 0$ we have: $$\sum_{n\geq 0}\frac{1}{a_n\cdot a_{n+1}\cdot\ldots\cdot a_{n+7}}=\frac{1}{7ab(a+b)(2a+b)(3a+b)(4a+b)(5a+b)(6a+b)}.\tag{4}$$
• I apologize: there was a typo. What I meant was $$\sum_{n=0}^{\infty} \frac{1}{a_{n}a_{n+1}\cdot\cdot\cdot a_{n+7}}$$. – mathlearner May 20 '15 at 11:45
• @mathlearner: there is just a hidden induction there. Are you able to tackle the simpler cases $$\sum_{n\geq 0}\frac{1}{(an+b)(an+a+b)}$$ and $$\sum_{n\geq 0}\frac{1}{(an+b)(an+a+b)(an+2a+b)}$$? If so, just induct. – Jack D'Aurizio May 20 '15 at 12:22