if $1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}=\frac{a_{n}}{b_{n}}$ there are infinitely many postive integers $n$,such $b_{n}>b_{n+1}$ Let $n$ be positive integers,such
$$1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n}=\dfrac{a_{n}}{b_{n}}$$
where $(a_{n},b_{n})=1$, show that
 there are infinitely many positive integers $n$,such $b_{n}>b_{n+1}$
 A: Addendum: I hadn't see Martin R's comment.  Go read the link he provided, it's a bit longer but actually complete and correct unlike my suggestion below.
Here is my reasoning.  You can write your finite sequence with a common denominator.  I'll write product from $1$ to $n$ missing factor $k$ by $1\cdots \hat k \cdots n$.
\begin{equation}
1 + \frac{1}{2} + \dots + \frac{1}{n} = \frac{\sum_{k=1}^n 1 \cdots \hat k \cdots n }{n!}
\end{equation}
You can simplify that last quotient into $\frac{a_n}{b_n}$, meaning there is some $d$ such that
\begin{align}
\sum_{k=1}^n 1 \cdots \hat k \cdots n &= d \cdot a_n\\
n! &= d \cdot b_n
\end{align}
Now consider going up to $n+1$.  The quotient becomes
\begin{align}
1 + \frac{1}{2} + \dots + \frac{1}{n+1} &= \frac{\sum_{k=1}^{n+1} 1 \cdots \hat k \cdots (n+1) }{(n+1)!} \\
  &=\frac{\big(\sum_{k=1}^n 1 \cdots \hat k \cdots n \big)(n+1) + n!}{(n+1)!}\\
  &= \frac{a_n\cdot d \cdot(n+1) + b_n \cdot d}{ b_n \cdot d \cdot (n+1)}
  = \frac{d \cdot (a_n\cdot(n+1) + b_n)}{ d \cdot (b_n \cdot (n+1))}
\end{align}
Thus,
\begin{equation}
  \frac{a_{n+1}}{b_{n+1}}
  = \frac{a_n\cdot(n+1) + b_n}{ b_n \cdot (n+1)}
\end{equation}
Now I'll try to to find a prime $p$ such that $p$ divides the denominator $b_n \cdot (n+1)$ and the numerator.  My intuition is that if I can build one, I can build many, hopefully enough to prove the claim.  To build one, $p$ must divide $b_n$ and $a_n\cdot(n+1)$
Let n = $2p-1$ for some prime $p$.  Then $p|n!$ exactly once. However
\begin{equation}
p \nmid \sum_{k=1}^n 1 \cdots \hat k \cdots n,
\end{equation}
Since $p | 1\cdots \hat k \cdots n$ for every $k \ne p$ yet $p \nmid 1\cdots \hat p \cdots n$.  Since $\sum_{k=1}^n 1 \cdots \hat k \cdots n = a_n \cdot d$, that implies $p \nmid d$.
But since $p|n!=d \cdot b_n$, that implies that $p | b_n$.  Obviously $p|n+1$ since $n+1 = 2\cdot p$.  Thus $\frac{a_n\cdot(n+1) + b_n}{ b_n \cdot (n+1)}$ can be simplified by $p$.
Now by choosing $n = \left(\prod_i p_i^{e_i}\right) - 1$, you should be able make a similar construction that ensure the simplification is large enough that ${b_{n+1}}<b_n$.  The critical point will probably where I proved $p \nmid \sum_{k=1}^n 1 \cdots \hat k \cdots n$, you'll have to choose your exponents $e_i$ very, very carefully to make $p_i^{e_i} \nmid \sum_{k=1}^n 1 \cdots \hat k \cdots n$ work.  I didn't try the computation so far, so my answer is not not definitive and should be taken with a grain of salt; maybe it doesn't work.  I hope it helps to guide you on how to solve your problem anyway.
