Sum of reciprocal of the terms of an arithmetic progression:

Let aa and dd be two coprime integers. (The great common divisor between aa and dd is 11). Prove that: (nN):1a+1a+d+1a+2d++1a+ndZ(\forall n\in\mathbb{N}^*):\frac{1}{a}+\frac{1}{a+d}+\frac{1}{a+2d}+\cdots+\frac{1}{a+nd}\notin\mathbb{Z}.

#Proofs #MathProblem #Math

Note by Mountassir Farid
7 years, 9 months ago

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3 votes

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Comments

Why is gcd(a,d)=1\gcd(a,d)=1? If this holds, then it also holds for a,da,d with gcd(a,d)>1\gcd(a,d)>1.

Tim Vermeulen - 7 years, 9 months ago

You could edit title as Sum of an Harmonic Progression....

Rahul Nahata - 7 years, 9 months ago

What about a=1a=1, d=1d=1, and n=n=\infty? Or is that not allowed? EDIT: would Z\infty\in\mathbb{Z}?

Daniel Liu - 7 years, 9 months ago

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∉N    ∉Z. \infty \not\in \mathbb{N} \implies \infty \not\in \mathbb{Z}.

Tim Vermeulen - 7 years, 9 months ago
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