# How to prove $\sum_{i=1}^{n-1}\frac{1}{\operatorname{lcm}(a_i,a_{i+1})}\lt1$ where $a_i\in\mathbb N$ and $a_i\lt a_{i+1}$?

Let $a_1,a_2,\ldots ,a_n\in\mathbb N$ and $a_1\lt a_2\lt\cdots\lt a_n$. Then how to prove $$\sum_{i=1}^{n-1}\frac{1}{\operatorname{lcm}(a_i,a_{i+1})}\lt1$$

By the way, a stronger statement is true: $\sum_{i=1}^{n-1} 1/LCM(a_i, a_{i+1}) \leq 1-1/2^n$. This one is a nice challenge. – David Speyer Mar 31 '13 at 20:12
Notice that $$S:=\sum_{i=1}^{\infty}\frac{1}{\text{lcm}\left(a_{i},a_{i+1}\right)}$$ $$=\sum_{i=1}^{\infty}\frac{\gcd(a_{i},a_{i+1})}{a_{i}a_{i+1}}$$ $$=\sum_{i=1}^{\infty}\frac{\gcd(a_{i},a_{i+1})}{a_{i+1}-a_{i}}\left(\frac{1}{a_{i}}-\frac{1}{a_{i+1}}\right).$$ 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 $1$.
Interpreting your last line, do you mean that since in the finite case we stop at $n-1$ and not infinity, then the sum comes out to $1/a_1-1/a_n$, and this is clearly less than 1? – Coffee_Table Mar 31 '13 at 20:23