A finite sum of fractions

Algebra Level 3

S = 2 − 1 2 + 1 3 − 1 4 + . . . + 1 2017 − 1 2018 + 1 2019 T = 1 1010 + 1 1011 + 1 1012 + . . . + 1 2019 \begin{aligned} S & = 2-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2017}-\dfrac{1}{2018}+\dfrac{1}{2019} \\ T & = \dfrac{1}{1010}+\dfrac{1}{1011}+\dfrac{1}{1012}+...+\dfrac{1}{2019} \end{aligned}

For S S and T T as defined above, calculate ( S − T ) 2019 (S-T)^{2019} .


The answer is 1.

Tin Le by Tin Le , 15, Vietnam

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2 solutions

Hugh Sir
Mar 12, 2019

S − T = [ 2 − 1 2 + 1 3 − 1 4 + . . . + 1 2017 − 1 2018 + 1 2019 ] − [ 1 1010 + 1 1011 + 1 1012 + . . . + 1 2019 ] S-T = [2-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2017}-\dfrac{1}{2018}+\dfrac{1}{2019}] - [\dfrac{1}{1010}+\dfrac{1}{1011}+\dfrac{1}{1012}+...+\dfrac{1}{2019}]

S − T = 1 + [ 1 + 1 2 + 1 3 + 1 4 + . . . + 1 2019 ] − 2 [ 1 2 + 1 4 + 1 6 + . . . + 1 2018 ] − [ 1 1010 + 1 1011 + 1 1012 + . . . + 1 2019 ] S-T = 1+[1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2019}]-2[\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{2018}]-[\dfrac{1}{1010}+\dfrac{1}{1011}+\dfrac{1}{1012}+...+\dfrac{1}{2019}]

S − T = 1 + [ 1 + 1 2 + 1 3 + 1 4 + . . . + 1 2019 ] − [ 1 + 1 2 + 1 3 + 1 4 + . . . + 1 1009 ] − [ 1 1010 + 1 1011 + 1 1012 + . . . + 1 2019 ] S-T = 1+[1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2019}]-[1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{1009}]-[\dfrac{1}{1010}+\dfrac{1}{1011}+\dfrac{1}{1012}+...+\dfrac{1}{2019}]

S − T = 1 + [ 1 + 1 2 + 1 3 + 1 4 + . . . + 1 2019 ] − [ 1 + 1 2 + 1 3 + 1 4 + . . . + 1 2019 ] S-T = 1+[1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2019}]-[1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2019}]

S − T = 1 S-T = 1

( S − T ) 2019 = 1 2019 = 1 (S-T)^{2019} = 1^{2019} = 1

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Chew-Seong Cheong
Mar 12, 2019

Consider the following:

S = 2 − 1 2 + 1 3 − 1 4 + ⋯ + 1 2017 − 1 2018 + 1 2019 = 1 + ( 1 + 1 2 + 1 3 + ⋯ + 1 2019 ) − 2 ( 1 2 + 1 4 + 1 6 + ⋯ + 1 2018 ) = 1 + ( 1 + 1 2 + 1 3 + ⋯ + 1 2019 ) − ( 1 + 1 2 + 1 3 + ⋯ + 1 1009 ) = 1 + H 2019 − H 1009 where H n is the n t h harmonic number. = 1 + 1 1010 + 1 1011 + 1 1012 + ⋯ + 1 2019 = 1 + T \begin{aligned} S & = 2 - \frac 12 + \frac 13 - \frac 14 + \cdots + \frac 1{2017} -\frac 1{2018} + \frac 1{2019} \\ & = 1+ \left(1 + \frac 12 + \frac 13 + \cdots + \frac 1{2019}\right) - 2\left(\frac 12 + \frac 14 + \frac 16 + \cdots + \frac 1{2018} \right) \\ & = 1+ \left(1 + \frac 12 + \frac 13 + \cdots + \frac 1{2019}\right) - \left(1+\frac 12 + \frac 13 + \cdots + \frac 1{1009} \right) \\ & = 1 + H_{2019} - H_{1009} & \small \color{#3D99F6} \text{where }H_n \text{ is the }n^{th} \text{ harmonic number.} \\ & = 1 + \frac 1{1010} + \frac 1{1011} + \frac 1{1012} + \cdots + \frac 1{2019} \\ & = 1 + T \end{aligned}

Therefore, ( S − T ) 2019 = 1 2019 = 1 (S-T)^{2019} = 1^{2019} = \boxed 1 .

Reference: Harmonic number

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