Inequality deduced from relatively prime numbers. If $a_n \text{ and }b_n$ are relatively prime for all $n$ and
$$\frac{a_n}{b_n}=\frac{1}{n}+\frac{1}{n(n+1)}+\frac{1}{n(n+1)(n+2)}+\cdots$$
Deduce that 
$$b_n\geq b_{n+1}$$
CURRENT THOUGHTS
I can show that 
$$\frac{a_{n+1}}{b_{n+1}}=\frac{na_n-b_n}{b_n}$$
and making $b_n$ the subject
$$b_n=(\frac{na_n}{a_{n+1}+b_{n+1}})b_{n+1}$$
so it would suffice to show that 
$$na_n\geq a_{n+1}+b_{n+1}$$
which would appear to be true in general as $n$ gets large. But other than this, I am unsure how to proceed? 
 A: I believe this statement naturally follows from the fact that $a_n$ and $b_n$ are defined to be coprime for all $n$. 
i.e.
If 
$$\frac{a_{n+1}}{b_{n+1}}=\frac{na_n-b_n}{b_n}$$
Then we must have that $b_n \geq b_{n+1}$ otherwise $a_{n+1}$ and $b_{n+1}$ wouldn't be coprime. 
Hope this helps! 
If I am not mistaken, the next steps in this proof will be to conclude that 
$$\frac{a_n}{b_n}>\frac{a_{n+1}}{b_{n+1}}\Rightarrow a_{n+1}<a_n$$
And we obtain a contradiction by 
$$a_1>a_2>a_3>\cdots >0$$
Since it is not possible to have a infinite decreasing sequence of positive integers? 
Very nice!
A: There is a simple straightforward proof that for each $n$ $$S(n)=\sum_{i=n}^\infty\prod_{j=n}^i \frac 1j$$ is irrational. Indeed, suppose to the contrary that $S(n)=\frac {a_n}{b_n}$ where $a_n$ and $b_n$ are integers. Put $m=n+b_n$. Among $b_n$ consecutive numbers $n, n+1,\dots, m-1$ exists one divisible by $b_n$. Then $$T(n)=n(n+1)\cdot\cdots\cdot (m-1)S(n)$$ is an integer. On the other hand, $T(n)$ is a sum of an integer and $S(m)$. But 
$$0<S(m)<\sum_{i=m}^\infty m^{m-i-1}=\frac 1m\cdot\frac{1}{1-\frac 1{m}}=\frac1{m-1}<1,$$
a contradiction.
