Partial zeta

Calculus Level 5

( n = 1 500 2 1 n ) ζ ( 1 2 ) \displaystyle \left\lfloor \left( \sum _{ n=1 }^{ { 500 }^{ 2 } }{ \frac { 1 }{ \sqrt { n } } } \right) -\zeta \left( \frac { 1 }{ 2 } \right) \right\rfloor

Determine the value of the expression above.

Image Credit: Wikimedia Geek3 .


The answer is 1000.

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1 solution

Aman Rajput
Jun 29, 2015

a = n = 1 250000 1 n = 998.541.. a = \sum_{n=1}^{250000} \frac{1}{\sqrt n} = 998.541..

b = ζ ( 1 / 2 ) = 1.4603.... b = \zeta(1/2) = -1.4603....

a b = 1000 \boxed {\lfloor a-b \rfloor = 1000}

But how did you find the sum? As a helpful hint, this is what I used:

( n = 1 x 2 1 n ) 2 x + ζ ( 1 2 ) \displaystyle \left( \sum _{ n=1 }^{ { x }^{ 2 } }{ \frac { 1 }{ \sqrt { n } } } \right) \approx 2x+\zeta \left( \frac { 1 }{ 2 } \right) as x goes to infinity.

Dylan Pentland - 5 years, 11 months ago

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How do you prove that?

Daniel Liu - 5 years, 11 months ago

That's an interesting relationship.

James Wilson - 3 years, 5 months ago

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