Joy of Mathematics

Double series with least common multiple

October 9, 2022October 9, 2022 Tolaso

Let $\mathrm{lcm}$ denote the least common multiple. Prove that

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

for all $s>1$ .

Solution

For each $M \in \mathbb{N}^+$ of the form $M =p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k}$ the number of solutions of $\mathrm{lcm} (n , m) = M$ is given by $(2a_1 + 1 ) (2a_2 + 1) \cdots (2a_k +1)$ . It follows that the series is equal to

$\sum_{M \geq 1} \frac{1}{M^s} \prod_{p \mid M} \left ( 2 \nu_p(M) + 1 \right )$

and since the function $M \mapsto \prod_{p \mid M} \left ( 2 \nu_p(M) + 1 \right )$ is multiplicative we have successively:

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

On a prime summation

November 13, 2020 Tolaso

Let $p_n$ denote the $n$ – th prime. Evaluate the sum

$\mathcal{S} = \sum_{n=1}^{\infty} \frac{ \log p_n}{p_n^2 -1}$

Sigma divisor sum

October 27, 2020 Tolaso

Let $\sigma$ denote the divisor function. Prove that

$\lim_{n \rightarrow +\infty} \frac{1}{n^2} \sum_{k=1}^{n} \sigma(k) = \frac{\pi^2}{12}$

Solution

We have successively:

$\begin{align*} \frac{1}{n^2} \sum_{k\leq n}\sigma(k) &= \frac {1}{n^2}\sum_{k \leq n}\sum_{q \mid k}q \\ &= \frac {1}{n^2}\sum_{q,d \mid qd \leq n}q \\ &= \frac {1}{n^2} \sum_{d\leq n} \sum_{q \leq n/d}q \\ &= \frac {1}{2n^2}\sum_{d \leq n} \left ( \frac {n^2}{d^2}+ \mathcal{O} \left ( \frac {n}{d}\right) \right ) \\ &= \frac{1}{2} \sum_{d \leq n} \frac{1}{d^2} + \mathcal{O} \left ( \frac{1}{n} \sum_{d \mid n} \frac{1}{d} \right ) \\ & = \frac{1}{2} \sum_{d \leq n} \frac{1}{d^2} + \mathcal{O} \left ( \frac{\ln n}{n} \right ) \\ & \rightarrow \frac{\pi^2}{12} \end{align*}$