33. [20] Compute
$$
\sum_{n=2009}^{\infty} \frac{1}{\binom{n}{2009}}
$$

Note that $\binom{n}{k}$ is defined as $\frac{n!}{k!(n-k)!}$.
Answer: $\qquad$ Observe that
$$
\begin{aligned}
\frac{k+1}{k}\left(\frac{1}{\binom{n-1}{k}}-\frac{1}{\binom{n}{k}}\right) & =\frac{k+1}{k} \frac{\binom{n}{k}-\binom{n-1}{k}}{\binom{n}{k}\binom{n-1}{k}} \\
& =\frac{k+1}{k} \frac{\binom{n-1}{k-1}}{\binom{n}{k}\binom{n-1}{k}} \\
& =\frac{k+1}{k} \frac{(n-1)!k!k!(n-k-1)!(n-k)!}{n!(n-1)!(k-1)!(n-k)!} \\
& =\frac{k+1}{k} \frac{k \cdot k!(n-k-1)!}{n!} \\
& =\frac{(k+1)!(n-k-1)!}{n!} \\
& =\frac{1}{\binom{n}{k+1}}
\end{aligned}
$$

Now apply this with $k=2008$ and sum across all $n$ from 2009 to $\infty$. We get
$$
\sum_{n=2009}^{\infty} \frac{1}{\binom{n}{2009}}=\frac{2009}{2008} \sum_{n=2009}^{\infty} \frac{1}{\binom{n-1}{2008}}-\frac{1}{\binom{n}{2008}} .
$$

All terms from the sum on the right-hand-side cancel, except for the initial $\frac{1}{\binom{2008}{2008}}$, which is equal to 1 , so we get $\sum_{n=2009}^{\infty} \frac{1}{\binom{n}{2009}}=\frac{2009}{2008}$.