13. Let $a_{1}=1, a_{2}=2$ and for all $n \geq 2, a_{n+1}=\frac{2 n}{n+1} a_{n}-\frac{n-1}{n+1} a_{n-1}$. It is known that $a_{n}>2+\frac{2009}{2010}$ for all $n \geq m$, where $m$ is a positive integer. Find the least value of $m$.
13. Answer: 4021
Rearranging the recurrence relation yields $a_{n+1}-a_{n}=\frac{n-1}{n+1}\left(a_{n}-a_{n-1}\right)$ for $n \geq 2$. Thus, for $n \geq 3$, we have
$$
\begin{aligned}
a_{n}-a_{n-1} & =\frac{n-2}{n}\left(a_{n-1}-a_{n-2}\right)=\frac{n-2}{n} \cdot \frac{n-3}{n-1}\left(a_{n-2}-a_{n-3}\right) \\
& =\cdots=\frac{n-2}{n} \cdot \frac{n-3}{n-1} \cdots \frac{2}{4} \cdot \frac{1}{3}\left(a_{2}-a_{1}\right) \\
& =\frac{2}{n(n-1)}=\frac{2}{n-1}-\frac{2}{n}
\end{aligned}
$$
for $n \geq 3$. Using the method of difference, we obtain $a_{n}=3-\frac{2}{n}$ for $n \geq 3$. Given that $a_{n}>2+\frac{2009}{2010}$, we have
$$
3-\frac{2}{n}>2+\frac{2009}{2010}=3-\frac{1}{2010},
$$
yielding $n>4020$. Hence the least value of $m$ is 4021 .