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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?

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  • $\begingroup$ Hint: If: $\frac{-b}{a}\notin \mathbb N$, then: $$\sum_{n=0}^\infty \frac{1}{a_na_{n+1}a_{n+7}}\le \sum_{n=1}^\infty \frac{1}{n^3}$$ $\endgroup$ – hamid kamali May 20 '15 at 11:34
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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: $$\sum_{n\geq 0}\frac{1}{a_n\cdot a_{n+1}\cdot\ldots\cdot a_{n+7}}=\frac{1}{7ab(a+b)(2a+b)(3a+b)(4a+b)(5a+b)(6a+b)}.\tag{4}$$

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  • $\begingroup$ I apologize: there was a typo. What I meant was $$\sum_{n=0}^{\infty} \frac{1}{a_{n}a_{n+1}\cdot\cdot\cdot a_{n+7}}$$. $\endgroup$ – mathlearner May 20 '15 at 11:45
  • $\begingroup$ @mathlearner: no problem, answer fixed. $\endgroup$ – Jack D'Aurizio May 20 '15 at 11:57
  • $\begingroup$ Thank you very much for your help. But combinatorics and fraction decompositions are really not my strong suits. I've been trying to understand your work for several minutes, but I'm getting nowhere. Would you mind adding some details and explanations in between the steps? $\endgroup$ – mathlearner May 20 '15 at 12:18
  • $\begingroup$ @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. $\endgroup$ – Jack D'Aurizio May 20 '15 at 12:22

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