Inverse arithmetic

Algebra Level 3

For any natural number N N is it true that there exists an strictly increasing sequence of N N positive integers in harmonic progression .

Clarification :

Inadequate information No Yes

Viki Zeta by Viki Zeta , 19, India

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1 solution

Viki Zeta
Oct 16, 2016

When AP is decreasing HP is increasing. Let d = -x a n = a + ( n − 1 ) d = a + ( n − 1 ) ( − x ) = ( a + x ) − n x a n + 1 = a + ( n + 1 − 1 ) d = a − n x ⟹ a n > a n + 1 ⟹ 1 a n + 1 > 1 a n ∴ As AP decreases HP increases, until N natural number. Here there is necessary to be a finite number of series because, when AP is decreasing one of the term will be ’0’ in some series, so HP = 1 0 which is indeterminate. \text{When AP is decreasing HP is increasing. } \\ \text{Let d = -x} \\ a_n = a + (n-1)d = a + (n-1)(-x) = (a + x) - nx \\ a_{n+1} = a + (n+1-1)d = a - nx \\ \implies a_n > a_{n+1} \\ \implies \dfrac{1}{a_{n+1}} > \dfrac{1}{a_n} \\ \therefore \text{As AP decreases HP increases, until N natural number. Here there is necessary to be a finite number of series because, } \\ \text{when AP is decreasing one of the term will be '0' in some series, so HP = } \dfrac{1}{0} \text{ which is indeterminate.}

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One possible answer for the skeptics: Let M > N M>N . Consider the sequence given by H n = 1 1 ( M − 1 ) ! − n − 1 M 1 ( M − 1 ) ! H_n=\frac{1}{\frac{1}{(M-1)!}-\frac{n-1}{M}\frac{1}{(M-1)!}} for 1 ≤ n ≤ N 1\leq n\leq N .

James Wilson - 3 years, 5 months ago

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