The series of reciprocals of the least common multiples of integers $X_1<X_2<\dots$ converges

Question

Let $X_1, X_2,\dots$ be a sequence of strictly increasing positive integers.

For each $n\ge1$, let $W_n$ be the least common multiple of the first $n$ terms $X_1, X_2,\dots, X_n$ of the sequence.

I need to prove the following statement:

The series $1/W_1+1/W_2+\dots+1/W_n\;(n\to\infty)$ is a convergent series.

I tried several ways, including the hint given below, but I have no luck to overcome this problem. I'd like to learn the methods that can be used to prove that such a series converges.

Answer 1

$\newcommand{\lcm}{\operatorname{lcm}}$ Note that for $n>1$ the least common multiple $W_n$ of natural numbers $1\leq X_1<X_2<\ldots<X_n$ fulfills \begin{align*} \lcm(X_{n-1},X_n) \leq W_n \end{align*}

The following theorem was originally conjectured by P. Erdös:

Theorem: Let $X_0,\ldots,X_n$ be integers satifying $1\leq X_0<X_1<\ldots<X_n$ then

\begin{align*} \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}\tag{1} \end{align*}

From (1) we obtain by noting that $W_1=\lcm(1,X_1)$

\begin{align*} \frac{1}{W_1}&+\frac{1}{W_2}+\frac{1}{W_3}+\cdots+\frac{1}{W_n}\\ &\leq \frac{1}{W_1}+\frac{1}{\lcm(X_1,X_2)}+\frac{1}{\lcm(X_2,X_3)}+\cdots+\frac{1}{\lcm(X_{n-1},X_n)}\\ &\leq 1-\frac{1}{2^{n}} \end{align*}

and OPs claim follows.

A proof of the theorem was given by D. Borwein in the paper A sum of reciprocals of least common multiples.

Proof (D. Borwein): For $j=1,2,\ldots,n$ let $$S_j=\frac{1}{\lcm(X_0,X_1)}+\cdots+\frac{1}{\lcm(X_{j-1},X_j)}$$

then $\lcm(X_{j-1},X_j)=u_jX_{j-1}=v_jX_{j}$ where $u_j>v_j\geq 1$. Hence \begin{align*} \frac{1}{\lcm(X_{j-1},X_j)}\leq \frac{1}{X_j}\tag{2} \end{align*} It follows from (2) that \begin{align*} \frac{1}{\lcm(X_{j-1},X_j)}\leq(u_j-v_j)\frac{1}{\lcm(X_{j-1},X_j)}=\frac{1}{X_{j-1}}-\frac{1}{X_j}\tag{3} \end{align*} We obtain from (3) \begin{align*} S_j\leq\frac{1}{X_0}-\frac{1}{X_j}\tag{4} \end{align*}

Now we show (1) is valid by considering all possible conditions on $X_0,X_1,\ldots,X_n$

Case 1: $X_n\leq 2^n$. It follows from (4),

\begin{align*} S_n\leq 1 - \frac{1}{X_n}\leq 1-\frac{1}{2^n} \end{align*}

Case 2: $X_j>2^j$ for $1\leq j \leq n$. We obtain by (2) \begin{align*} S_n\leq\frac{1}{X_1}+\frac{1}{X_2}+\cdots+\frac{1}{X_n}<\frac{1}{2}+\frac{1}{2^2}+\cdots+\frac{1}{2^n}=1-\frac{1}{2^n} \end{align*}

Case 3: $X_k\leq2^k$ for some positive integer $k<n$ and $X_j>2^j$ for $k+1\leq j\leq n$. Then, by (2) and (4) \begin{align*} S_n&=S_k+\frac{1}{\lcm(X_{k},X_{k+1})}+\cdots+\frac{1}{\lcm(X_{n-1},X_n)}\\ &<1-\frac{1}{2^k}+\frac{1}{2^{k+1}}+\cdots+\frac{1}{2^n}\\ &=1-\frac{1}{2^n} \end{align*}

We conclude (1) holds in all cases and the theorem follows.

Answer 2

Hint:

If $W_{n+1}>W_n$ then $W_{n+1}$ has at least one more prime factor than $W_n$, so $W_{n+1}\ge 2W_n$

Answer 3

Since each of $X_1,\dots,X_n$ are distinct and divide $W_n$ we have that $$n\leq d(W_n)$$ where $d(n)$ is the number of divisors function. Using the classical result of Severin Wigert that $$\limsup_{n\rightarrow\infty}\frac{\log d(n)}{\log n/\log\log n}=\log2,$$ which is obtained by simply considering integers of the form $\prod_{p\leq x}p$, [See Montgomery and Vaughn chapter 2] we have that $n$ sufficiently large, $$\log n\leq\log d(W_{n})\leq\frac{\log W_{n}}{\log\log W_{n}},$$ and so $$W_{n}\geq n^{\log\log n}$$ when $n$ is sufficiently large. This implies that $$\sum_{n}\frac{1}{W_{n}}$$ converges as desired.

Answer 4

Let $W (n)$ be 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 $\prod_{j = 0}^{k} m_j > 2^k$ 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

$$\sum_{n = 1}^{\infty} \frac {1} {W (n)} < \frac {1} {W_1} \sum_{n = 1}^{\infty} \frac {n} {2^n - 1} = \frac {2.74403...} {W_1}$$

So the sum of reciprocals of $W (n)$ converges. Then, by Cesaro summation, we see that $$\sum_{n = 1}^{\infty} \frac {1} {W_n}$$ also converges.