# Hint: Zeta is involved!

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

Find a closed form for the above summation for $a>4$.

This is a part of the set Formidable Series and Integrals Note by Hamza A
5 years, 2 months ago

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## Comments

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we 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 $\displaystyle\sum _{ n=1 }^{ \infty }{ \sum _{ k=1 }^{ \infty }{ \dfrac { 1 }{ \left\lfloor \sqrt { n+k } \right\rfloor ^{ a } } } } =2\zeta(a-3)+3\zeta(a-2)-\zeta(a-1)-\zeta(a)$

- 5 years, 2 months ago

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coool ... :)

- 5 years, 2 months ago

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May be $\large 2a\zeta(a-1)$

- 5 years, 2 months ago

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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?

- 5 years, 2 months ago

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Post if you can :)

- 5 years, 2 months ago

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