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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

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


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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1 Comment

  1. 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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