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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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What have you tried to do in this direction? For instance, the first thing that you could try is this: if $n = 0$, then $\{kn + j\}_{1\leq j \leq k} = \{1, 2, \dots, k\}$. When $n = 1$, you obviously get $\{kn + j\}_{1\leq j\leq k} = \{k + 1, k + 2, \dots, k+k = 2k\}$. Hence $\{kn + j\}_{1\leq j\leq k, n = 0, 1} = \{1,2,\dots,k, k+1,\dots 2k\}$. Notice a pattern? Can you generalize this and write it down formally? –  William Feb 18 '12 at 1:07
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Set $k=3$ and write out maybe ten terms in both sums. See it? –  anon Feb 18 '12 at 1:13
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2 Answers 2

up vote 1 down vote accepted

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 $m = n k + j$.

enter image description here

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