Cómo demostrar a $\sum_{i=1}^{n-1}\frac{1}{\operatorname{lcm}(a_i,a_{i+1})}\lt1$ donde$a_i\in\mathbb N$$a_i\lt a_{i+1}$?

Question

Deje $a_1,a_2,\ldots ,a_n\in\mathbb N$$a_1\lt a_2\lt\cdots\lt a_n$. Entonces, ¿cómo demostrar $$\sum_{i=1}^{n-1}\frac{1}{\operatorname{lcm}(a_i,a_{i+1})}\lt1$$

Gracias de antemano

Answer 1

Observe que $$S:=\sum_{i=1}^{\infty}\frac{1}{\text{lcm}\left(a_{i},a_{i+1}\right)}$$ $$=\sum_{i=1}^{\infty}\frac{\gcd(a_{i},a_{i+1})}{a_{i}a_{i+1}}$$ $$=\sum_{i=1}^{\infty}\frac{\gcd(a_{i},a_{i+1})}{a_{i+1}-a_{i}}\left(\frac{1}{a_{i}}-\frac{1}{a_{i+1}}\right).$$ Since $\gcd(a_{i},a_{i+1})=\gcd(a_{i},a_{i+1}-a_{i})\leq a_{i+1}-a_{i},$ and $a_{i+1}>a_i$, the coefficient $\frac{\gcd(a_{i},a_{i+1})}{a_{i+1}-a_{i}}$ is $\leq 1$. Thus we may bound the series above by $$\sum_{i=1}^{\infty}\left(\frac{1}{a_{i}}-\frac{1}{a_{i+1}}\right)=\frac{1}{a_{1}}\leq1. $$ It follows that the finite sum is always strictly less than $1$.