Problem for the next year!

Algebra Level 4

1 1 − 1 2 + 1 3 − 1 4 + … + 1 2015 − 1 2016 1 1009 + 1 1010 + 1 1011 + … + 1 2015 + 1 2016 = ? \large{\dfrac{ \displaystyle \frac{1}{1}-\frac{1}{2}+\frac{1}{3} - \frac{1}{4} + \ldots + \frac{1}{2015}-\frac{1}{2016} }{\displaystyle \frac{1}{1009}+\frac{1}{1010} + \frac{1}{1011} + \ldots + \frac{1}{2015}+\frac{1}{2016} } = \ ? }


The answer is 1.000.

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Satyajit Mohanty by Satyajit Mohanty , 24, India

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2 solutions

Akshat Sharda
Sep 15, 2015

= 1 − 1 2 + 1 3 − 1 4 + . . . − 1 2016 =1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...-\frac{1}{2016}

= 1 + 1 2 + 1 3 + 1 4 + . . . + 1 2016 − 2 ( 1 2 + 1 4 + 1 6 + . . . + 1 2016 ) =1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}-2(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016})

= 1 + 1 2 + 1 3 + 1 4 + . . . + 1 2016 − ( 1 + 1 2 + 1 3 + . . . + 1 1008 ) =1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}-(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1008})

= 1 1009 + 1 1010 + . . . + 1 2016 =\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}

so the answer is 1 . \boxed{1}.

32 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice solution, Upvoted!!!!!!!!!!!!!!!!!!!!

Department 8 - 5 years, 9 months ago

CORRECTION:See the comments!

Harmonic sum can be written as H n = l n ( n ) + γ + a l p h a 0 Hn=ln(n)+\gamma+alpha0

So denominator becomes equal to l n 2 ln 2 and numerator becomes equal to l n 2016 − l n 1008 ln2016-ln1008

So the result is 1 1 .

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Well... I disagree with your solution. It's not accurate. Please refer Akshat Sharda's solution for the correct approach.

Satyajit Mohanty - 5 years, 9 months ago

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