Compare a a and b b

Algebra Level 5

a = 1 − 1 2 + 1 3 − 1 4 + 1 5 − 1 6 + ⋯ + 1 1999 − 1 2000 b = 1 1000 + 1 1001 + 1 1002 + ⋯ + 1 2000 \begin{aligned} a&=& 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots+ \frac{1}{1999}-\frac{1}{2000} \\ b&= &\frac{1}{1000} + \frac{1}{1001} + \frac{1}{1002} + \cdots + \frac{1}{2000} \end{aligned}

Which of the following is true?

a > b a > b a < b a < b a = b a = b a a and b b can't be compared

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Adrian Neacșu by Adrian Neacșu , 33, Romania

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3 solutions

Chew-Seong Cheong
Sep 10, 2014

a = 1 − 1 2 + 1 3 − 1 4 + 1 5 − 1 6 . . . + 1 1999 − 1 2000 a = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} ... + \frac{1}{1999} - \frac{1}{2000}

= 1 + ( 1 2 − 1 ) + 1 3 + ( 1 4 − 1 2 ) + 1 5 + ( 1 6 − 1 3 ) . . . + 1 1999 + ( 1 2000 − 1 1000 ) \quad = 1 + (\frac{1}{2} - 1) + \frac{1}{3} + (\frac{1}{4} - \frac{1}{2} ) + \frac{1}{5} + (\frac{1}{6} - \frac{1}{3} ) ... + \frac{1}{1999 }+ (\frac{1}{2000} - \frac{1}{1000} )

= 1 + 1 2 + 1 3 . . . + 1 2000 − 1 − 1 2 − 1 3 . . . − 1 1000 \quad = 1 + \frac{1}{2} + \frac{1}{3} ... + \frac{1}{2000} - 1 - \frac{1}{2} - \frac{1}{3} ... - \frac{1}{1000}

⇒ a = ∑ n = 1 2000 1 n − ∑ n = 1 1000 1 n \Rightarrow a =\sum _{ n=1 }^{ 2000 }{ \frac { 1 }{ n } } - \sum _{ n=1 }^{ 1000 }{ \frac { 1 }{ n } }

b = 1 1000 + 1 1001 + 1 1002 . . . + 1 2000 b = \frac{1}{1000} + \frac{1}{1001} + \frac{1}{1002} ... + \frac{1}{2000}

⇒ b = ∑ n = 1 2000 1 n − ∑ n = 1 999 1 n \Rightarrow b =\sum _{ n=1 }^{ 2000 }{ \frac { 1 }{ n } } - \sum _{ n=1 }^{ 999 }{ \frac { 1 }{ n } }

It is noticed that b − a = 1 1000 ⇒ a < b b - a = \frac{1}{1000} \quad \Rightarrow \boxed { a < b }

65 Helpful 0 Interesting 0 Brilliant 0 Confused

Excellent!

Lu Chee Ket - 5 years, 6 months ago

Love this solution!!!

B.S.Bharath Sai Guhan - 6 years, 8 months ago

careless.. Is this valid?

Let c = 1 + 1 2 + 1 3 + . . . + 1 1000 c = 1 + \frac{1}{2} +\frac{1}{3} + ... + \frac{1}{1000}

It is exactly that a + b + c − 1 1000 = 2 ( 1 + 1 3 + 1 1 5 + . . . + 1 1999 ) a+b+c - \frac{1}{1000} = 2(1 + \frac{1}{3} + 1\frac{1}{5} + ... + \frac{1}{1999}) a + b + c − 1 1000 = 2 ( b + c − 1 1000 − 1 2 − 1 4 − . . . − 1 2000 ) a+b+c- \frac{1}{1000} = 2(b+c - \frac{1}{1000} - \frac{1}{2} - \frac{1}{4} - ... - \frac{1}{2000}) a + b + c − 1 1000 = 2 ( b + c − 1 1000 − 1 2 ( 1 + . . . + 1 1000 ) a+b+c - \frac{1}{1000} = 2(b+c - \frac{1}{1000} -\frac{1}{2} (1 + ... + \frac{1}{1000}) a + b + c − 1 1000 = 2 ( b + c − 1 1000 − c 2 ) a+b+c - \frac{1}{1000} = 2(b+c - \frac{1}{1000} - \frac{c}{2}) a + b + c − 1 1000 = 2 b + c − 2 1000 a+b+c - \frac{1}{1000} = 2b+c-\frac{2}{1000} a + b + c = 2 b + c − 1 1000 a+b+c = 2b+c- \frac{1}{1000} a = b − 1 1000 < b a=b-\frac{1}{1000} < b

It concludes that a < b a<b

I made a very stupid mistake during subtracting

Figel Ilham - 5 years, 6 months ago

Beautiful solution

Pranav Rao - 5 years, 6 months ago

Did the same way.

Anupam Nayak - 5 years, 5 months ago

I missed the term 1 1000 \frac{1}{1000} in B. >__<

Jared Low - 6 years, 5 months ago
AviRal VerMa
Sep 9, 2014

a= 1-1/2+1/3.....-1/2000

a= (1+1/3+......1/1999) - ( 1/2 +.......+1/2000)

a=(1 + 1/2 +1/3.......+1/2000) -( 1/2 +.......+1/2000)- ( 1/2 +.......+1/2000)

a= (1 + 1/2 +1/3.......+1/2000) -2 x ( 1/2 +.......+1/2000)

a= (1 + 1/2 +1/3.......+1/2000) -2 /2 ( 1/1 +.......+1/1000) [taking two common from second bracket]

a=(1 + 1/2 +1/3.......+1/2000) - ( 1/1 +.......+1/1000)

a=1/1000 +1/1001 .....1/2000=b

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Actually, a a is equal to 1 1001 + 1 1002 + 1 1003 + ⋯ + 1 2000 . \frac{1}{1001} + \frac{1}{1002} + \frac{1}{1003} + \dots + \frac{1}{2000}. If this is the intended expression for b b , then the problem should be fixed.

Jon Haussmann - 6 years, 9 months ago

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Thanks for pointing that out. I've updated the answer to a < b a < b , given the phrasing.

In future, if you spot any errors with a problem, you can “report” it by selecting the “dot dot dot” menu in the lower right corner. You will get a more timely response that way.

Calvin Lin Staff - 6 years, 9 months ago

@Adrian Neacșu I've updated the expression of a a to make the 2nd last term 1 1999 \frac{1}{1999} instead of 1 2000 \frac{1}{ 2000} .

Calvin Lin Staff - 6 years, 9 months ago

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Thanks but why does it say that a < b a < b

Adrian Neacșu - 6 years, 9 months ago

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See Jon's comment above.

Calvin Lin Staff - 6 years, 9 months ago

Ouch pressed the wrong button but i have the same solution as yours :)

Rindell Mabunga - 6 years, 9 months ago

In order to b − a = 1 1000 b-a=\frac{1}{1000} , the definition of b should be changed to b = 1 1000 + 1 1001 + 1 1002 + . . . + 1 2000 b=\frac { 1 }{ 1000 } +\frac { 1 }{ 1001 } +\frac { 1 }{ 1002 } +...+\frac { 1 }{ 2000 } .

João Henrique Franco - 6 years, 9 months ago

I interpreted b b as: 1 1000 + 1 1001 + 1 1003 + 1 1006 . . . \frac { 1 }{ 1000 } +\frac { 1 }{ 1001 } +\frac { 1 }{ 1003 } +\frac { 1 }{ 1006 } ... Due to the problem showing b = 1 1000 + 1 1001 + 1 1003 . . . b=\frac { 1 }{ 1000 } +\frac { 1 }{ 1001 } +\frac { 1 }{ 1003 } ... instead of ∑ n = 0 1000 1 1000 + n \sum _{ n=0 }^{ 1000 }{ \frac { 1 }{ 1000+n } }

I feel this is unfair. Any thoughts?

Julian Poon - 6 years, 8 months ago

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Thanks for pointing it out. I have updated the phrasing of the problem.

@Adrian Neacșu Please take care when you write up a problem. Review it carefully especially if several people have submitted a dispute.

Calvin Lin Staff - 6 years, 8 months ago

Any body having a rational mind can do it without using pen and paper ! is't it easy and fun ? :)

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