×

# 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 Hummus A
1 year, 3 months ago

Sort by:

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

coool ... :) · 1 year, 3 months ago

May be $\large 2a\zeta(a-1)$ · 1 year, 3 months ago

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? · 1 year, 3 months ago

Post if you can :) · 1 year, 3 months ago

Comment deleted Apr 11, 2016

Comment deleted Apr 11, 2016

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. · 1 year, 3 months ago

You misunderstood this question. · 1 year, 3 months ago