22. A sequence $a_{1}, a_{2}, a_{3}, a_{4}, \cdots$, with $a_{1}=\frac{1}{2}$, is defined by
$$
a_{n}=2 a_{n} a_{n+1}+3 a_{n+1}
$$
for all $n=1,2,3, \cdots$. If $b_{n}=1+\frac{1}{a_{n}}$ for all $n=1,2,3, \cdots$, find the largest integer $m$ such that
$$
\sum_{k=1}^{n} \frac{1}{\log _{3} b_{k}}>\frac{m}{24}
$$
for all positive integer $n \geq 2$.
22. Answer: 13
Solution. Given that $a_{n}=2 a_{n} a_{n+1}+3 a_{n+1}$ we obtain $a_{n+1}=\frac{a_{n}}{2 a_{n}+3}$. Thus we have $\frac{1}{a_{n+1}}=2+\frac{3}{a_{n}}$. We thus have $\frac{1}{a_{n+1}}+1=3\left(1+\frac{1}{a_{n}}\right)$ for all $n=1,2,3, \ldots$. Letting $b_{n}=1+\frac{1}{a_{n}}$, it is clear that the sequence $\left\{b_{n}\right\}$ follows a geometric progression with first term $b_{1}=1+\frac{1}{a_{1}}=3$, and common ratio 3 . Thus, for $n=1,2,3, \cdots, b_{n}=1+\frac{1}{a_{n}}=3^{n}$ for $n=1,2,3, \cdots$.
Let $f(n)=\sum_{k=1}^{n} \frac{1}{n+\log _{3} b_{k}}=\sum_{k=1}^{n} \frac{1}{n+k}>\frac{m}{24}, n=2,3,4, \cdots$. It is clear that $f(n)$ is an increasing function since
$$
f(n+1)-f(n)=\frac{1}{n+1}>0 .
$$

Thus $f(n)$ is a strictly increasing sequence in $n$. Thus the minimum value of $f(n)$ occurs when $n=2$.
$$
f(2)=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}>\frac{m}{24},
$$
forcing $m<14$. Thus the largest value of integer $m$ is 13 .