\[\large \displaystyle\sum _{ n=1 }^{ \infty }{ \sum _{ k=1 }^{ \infty }{ \dfrac { 1 }{ \left\lfloor \sqrt { n+k } \right\rfloor ^{ a } } } } \]

Find a closed form for the above summation for \(a>4\).

This is a part of the set Formidable Series and Integrals

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## Comments

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TopNewestwe have \[\displaystyle\sum _{ n=1 }^{ \infty }{ \sum _{ k=1 }^{ \infty }{ \dfrac { 1 }{ \left\lfloor \sqrt { n+k } \right\rfloor ^{ a } } } } =\displaystyle\sum _{ i=1 }^{ \infty }{ \left( \displaystyle\sum _{ { i }^{ 2 }\le k+n<(i+1)^{ 2 } }{ \frac { 1 }{ \left\lfloor \sqrt { n+k } \right\rfloor ^{ a } } } \right) } =\displaystyle\sum _{ i=1 }^{ \infty }{ \frac { 1 }{ { i }^{ a } } \left( \sum _{ { i }^{ 2 }\le k+n<(i+1)^{ 2 } }{ 1 } \right) } \]

evaluating the inner sum we get \(2i^3+3i^2-i-1\)

so it follows that \[\displaystyle\sum _{ n=1 }^{ \infty }{ \sum _{ k=1 }^{ \infty }{ \dfrac { 1 }{ \left\lfloor \sqrt { n+k } \right\rfloor ^{ a } } } } =2\zeta(a-3)+3\zeta(a-2)-\zeta(a-1)-\zeta(a)\]

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coool ... :)

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May be \[\large 2a\zeta(a-1)\]

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i'm getting

\(2\zeta(a-3)+3\zeta(a-2)-\zeta(a-1)-\zeta(a)\)

do you want me to post how i got the answer?

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Post if you can :)

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Comment deleted Apr 11, 2016

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Comment deleted Apr 11, 2016

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No the n&k values are used to clarify the most simple values for a, an infinite series has up to the given below number, not every number possibly out there. I can argue this and be right.

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