A series with least common multiple

Let $\{X_n\}_{n \in \mathbb{N}}$ be a strictly increasing sequence of positive numbers. For all $n \geq 1$ denote as $W_n$ the least common multiple of the first $n$ terms $X_1, X_2, \dots, X_n$ of the sequence. Prove that , as $n \rightarrow +\infty$ , the following sum converges

$\mathcal{S} = \frac{1}{W_1} + \frac{1}{W_2} + \cdots + \frac{1}{W_n}$

Solution

This is a result due to Paul Erdös stating that if $X_1, X_2, \dots, X_n$ are natural numbers such that $1 \leq X_1 < X_2 < \cdots < X_n$ then

$\displaystyle \frac{1}{\lcm(X_0,X_1)}+\frac{1}{\lcm(X_1,X_2)}+\cdots+\frac{1}{\lcm(X_{n-1},X_n)}\leq 1-\frac{1}{2^n}$

and the original question follows since the sum we seek is less or equal to $\displaystyle 1 - \frac{1}{2^n}$ .

However, we are presenting another proof. Denote as $W(n)$ the average order of the numbers $W_n$ , i.e.,

$W (n) = \frac {1} {n} \sum_{k = 1}^{n} W_k.$

For any $k$ we have $W_{k + 1} = W_k \cdot m_k$ where $m_k$ is the product of primes not present in the factorization of $X_1, X_2, \cdots, X_k$ . Note that $m_k$ are squarefree integers. Note also that it may be an empty product, i.e., $m_k = 1$ . Then

$\sum_{k = 1}^{n} W_k = W_1 \cdot \sum_{k = 0}^{n - 1} \prod_{j = 0}^{k} m_j.$

It is easy to see (and show by induction) that $\prod \limits_{j = 0}^{k} m_j > 2^k$ so we have

$\sum_{k = 1}^{n} W_k > W_1 \cdot \sum_{k = 0}^{n - 1} 2^k = (2^n - 1) W_1.$

Hence, $W (n) > \frac {2^n - 1} {n} W_1.$ Consequently, we have

$\sum_{n = 1}^{\infty} \frac {1} {W (n)} < \frac {1} {W_1} \sum_{n = 1}^{\infty} \frac {n} {2^n - 1}$

So the sum of reciprocals of $W (n)$ converges. Then, by Cesàro summation, we see that

$\sum_{n = 1}^{\infty} \frac {1} {W_n}$

also converges.

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One thought on “A series with least common multiple”

John Khan says:

August 30, 2017 at 10:30 pm

Probably a familiar series for testing of convergence is:

Let $\{a_n\}_{n \in \mathbb{N}}$ be a a strictly increasing sequence of positive integers. Prove that the series $\displaystyle \sum_{n=1}^{\infty} \frac{1}{[a_n , a_{n+1}]}$ converges where $[ \; ]$ denotes the least common multiple.

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