A number theory problem by Wildan Bagus Wicaksono

1 + 1 n + 1 n 2 + 1 n 3 + = x y 1+\frac { 1 }{ n } +\frac { 1 }{ { n }^{ 2 } } +\frac { 1 }{ { n }^{ 3 } } +\cdots=\frac { x }{ y }

Let x , y x,y and n n be positive integers larger than 1.

Given that x x and y y are coprime. Find x y x-y .


The answer is 1.

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1 solution

Marta Reece
Jul 2, 2017

1 + 1 n + 1 n 2 + 1 n 3 + . . . 1+\frac { 1 }{ n } +\frac { 1 }{ { n }^{ 2 } } +\frac { 1 }{ { n }^{ 3 } } +... is a geometric series with first term a = 1 a=1 and common ratio r = 1 n r=\frac 1n .

The sum of the series is S = a 1 r = 1 1 1 n = n n 1 = x y S=\dfrac a{1-r}=\dfrac 1{1-\frac 1n}=\dfrac n{n-1}=\dfrac xy

The difference x y = n ( n 1 ) = 1 x-y=n-(n-1)=\boxed1

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