[m]S_{n}=\sum_{1}^{n}\frac{1}{(k+3)(k+4)}=\sum_{1}^{n}\frac{1}{k+3}-\sum_{1}^{n}\frac{1}{k+4}=[/m]
[m]=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{n+2}-\frac{1}{n+3}+\frac{1}{n+3}-\frac{1}{n+4}=\frac{1}{4}-\frac{1}{n+4}[/m]
По определению сумма ряда
S=lim_(n → ∞ )S_(n)=[m]\frac{1}{4}-0=\frac{1}{4}[/m]