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If Harmonic Sequence

$$ H_n=\left\{ {1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5},...} \right\} $$

How can i find an Arithmetic Sequence using the above sequence?

Actually,i have found one by inspection. $$ \left\{{\frac{1}{3},\frac{1}{4},\frac{1}{6}}\right\}\\ It\; is\; a\; Arithmetic\; Sequence\; with\; common\; difference=\;\frac{-1}{6} $$ However,i have no idea on finding the other arithemetic sequence with more term. Thank you.

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1 Answer 1

up vote 2 down vote accepted

Note that your example can be written over the common denominator 12 as $$\frac{4}{12}, \frac{3}{12}, \frac{2}{12}.$$ This suggests starting with a (decreasing) arithmetic progression of natural numbers, then finding common denominator, and turning it into fractions.

This likely would give long progressions.

EDIT: Try the sequence $$\frac{n-k}{n!}$$ for $k=0,1,2...;$ for example $n=5$ gives $$\frac{1}{24},\frac{1}{30},\frac{1}{40},\frac{1}{60},\frac{1}{120}.$$

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I am now able to find 6 term arithmetic progression with your hint of common divisor 12,but i still cannot find the higher term sequence –  Vulcan Dec 8 '12 at 13:23
    
Oh,I know how to do it now.If the common divisor is 12,then we have 4 term sequence since 1,2,3,4 divides 12.If i choose a number k>n and 1,2,3,...,n divides k.Therefore we could get a n term sequence.Anyway,thanks for helping –  Vulcan Dec 8 '12 at 13:31
    
Yes that works. See added remark under EDIT, using factorials. –  coffeemath Dec 8 '12 at 13:59
    
Thanks for yours help^^ –  Vulcan Dec 8 '12 at 14:06

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