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The Equality is that:

$\sum\limits_{n=0}^{\infty} \sum\limits_{j=1}^{k}\cfrac{1}{(kn+j)^s}=\sum\limits_{n=0}^{\infty} \cfrac{1}{(n+1)^s} $ , where s>1 .

how to show that is true?

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    Set $k=3$ and write out maybe ten terms in both sums. See it?2012-02-18

2 Answers 2

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this is true

for any $k \in \mathbb{N}$ take $I_{k,n}=[kn+1 , k(n+1)]$

you can see that $\mathbb{N}^*=\cup_{n \in \mathbb{N}}I_{k,n}$

as $\frac{1}{(n+1)^s}>0 , \forall n \in \mathbb{N} $

then $\sum\limits_{n\in \mathbb{N}^*} \cfrac{1}{n^s} = \sum\limits_{n\in \mathbb{N}} \sum\limits_{p\in I_{k,n}}\cfrac{1}{p^s}$

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Any number $m \in \mathbb{N}$ can be represented through quotient and remainder of division by $k$, as $m = n k + j$.

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