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 Post subject: Series with least common multiple.Posted: Sun Aug 27, 2017 6:06 am

Joined: Sat Nov 07, 2015 6:12 pm
Posts: 841
Location: Larisa
Let ${\rm lcm}$ denote the least common multiple . Prove that for all $s>1$ the following holds:

$$\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{{\rm lcm}^s(m, n)} = \frac{\zeta^3(s)}{\zeta(2s)}$$

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 Post subject: Re: Series with least common multiple.Posted: Sat May 26, 2018 1:11 pm

Joined: Sat Nov 14, 2015 6:32 am
Posts: 159
Location: Melbourne, Australia
An elementary approach would be as follows:

For all $M \in \mathbb{N}^+$ of the form $M=p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k}$ the number of solutions

$$\mathrm{lcm} (m, n) =M$$

is given by $(2a_1+1) (2a_2+1) \cdots (2a_k+1)$. It follows that the given series equals

$$\sum_{M =1}^{\infty} \frac{1}{M^s} \prod_{p| M} \left ( 2 \nu_p(M)+1 \right )$$

and since the function $\prod \limits_{p| M} \left ( 2 \nu_p(M)+1 \right )$ is clearly a multiplicative one , it follows by Euler's product that

\begin{align*}
\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{{\rm lcm}^s(m, n)} &= \prod_{p \in \mathbb{P}} \left ( 1 + \frac{3}{p^s} + \frac{5}{p^{2s}} + \frac{7}{p^{3s}} + \cdots \right ) \\
&= \prod_{p \in \mathbb{P}} \frac{p^s \left ( p^s+1 \right )}{\left ( p^s-1 \right )^2}\\
&=\prod_{p \in \mathbb{P}} \frac{1-\frac{1}{p^{2s}}}{\left ( 1 - \frac{1}{p^s} \right )^3} \\
&= \frac{\zeta^3(s)}{\zeta(2s)}
\end{align*}

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$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$

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