$\begin{aligned} & \text { Let } \quad S=\frac{1}{1 \cdot 3}+\frac{1}{2 \cdot 5}+\frac{1}{3 \cdot 7}+\frac{1}{4 \cdot 9}+\ldots \\ & \therefore \quad T_n=\frac{1}{n(2 n+1)} \\ & =\frac{1}{n}-\frac{2}{(2 n+1)} \\ & \Rightarrow \quad S=\sum_{n=1}^{\infty} T_n=\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{2}{2 n+1}\right) \\ & =\frac{1}{1}-\frac{2}{3}+\frac{1}{2}-\frac{2}{5}+\frac{1}{3}-\frac{2}{7}+\frac{1}{4}-\frac{2}{9}+\frac{1}{5}-\ldots \\ & =1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\ldots \\ & =1-\left(-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\ldots\right) \\ & =1-\left(-1+\log _e 2\right)=2-\log _e 2 \\ & \end{aligned}$