\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{n\left(n+2\right)}< \frac{2003}{2004}\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{n}+\frac{1}{n+2}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{n+2}\right)\)
\(=\frac{1}{2}\left(\frac{n+2}{n+2}-\frac{1}{n+2}\right)\)
\(=\frac{1}{2}.\frac{n+1}{n+2}\)
\(=\frac{n+1}{2\left(n+2\right)}< \frac{2003}{2004}\)
\(\Leftrightarrow\hept{\begin{cases}n+1< 2003\\2\left(n+2\right)< 2004\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}n< 2002\\\left(n+2\right)< 1002\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}n< 2002\\n< 1000\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}n+1=2002\\2\left(n+2\right)=1000\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}n=2001\\n=498\end{cases}}\)
