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theo mình 2 bài này phải áp dụng bđt |$a_{1}+a_{2}+a_{3}+...+a_{n}|$$\leq$$|a_{1}|+|a_{2}|+|a_{3}|+...+|a_{n}|$.câu 1:$u_{n}\leq |u_{n}|=|\frac{cos1}{1.2}+\frac{cos2}{2.3}+...+\frac{cosn}{n(n+1)}|\leq |\frac{cos1}{1.2}|+|\frac{cos2}{2.3}|+...+|\frac{cosn}{n(n+1)}|\\$$\leq \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{n(n+1)}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}=1-\frac{1}{n+1}\\<1\rightarrow u_{n}<1$câu 2:bằng phương pháp quy nạp ta cm được $\frac{1}{n(n+1)(n+2)}=\frac{1}{2}\left[ \frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)} {} \right]$tương tự câu 1,$u_{n}\leq |u_{n}|\leq \frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{n(n+1)(n+2)}$$=\frac{1}{2}\left[ \frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}{} \right]$$=\frac{1}{2}\left[ \frac{1}{2}-\frac{1}{(n+1)(n+2)}{} \right]$$=\frac{1}{4}-\frac{1}{2(n+1)(n+2)}<\frac{1}{4}\rightarrow u_{n}<\frac{1}{4}$.mong bạn đọc cho ý kiến!!!

theo mình 2 bài này phải áp dụng bđt |$a_{1}+a_{2}+a_{3}+...+a_{n}|$$\leq$$|a_{1}|+|a_{2}|+|a_{3}|+...+|a_{n}|$.câu 1:$u_{n}\leq |u_{n}|=|\frac{cos1}{1.2}+\frac{cos2}{2.3}+...+\frac{cosn}{n(n+1)}|\leq |\frac{cos1}{1.2}|+|\frac{cos2}{2.3}|+...+|\frac{cosn}{n(n+1)}|\\$$\leq \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{n(n+1)}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}=1-\frac{1}{n+1}\\<1\rightarrow u_{n}<1$câu 2:bằng phương pháp quy nạp ta cm được $\frac{1}{(n-2)(n-1)n}=\frac{1}{2}\left[ \frac{1}{(n-2)(n-1)}-\frac{1}{(n-1)n} {} \right]$ với n$\geq 3\in N$tương tự câu 1,$u_{n}\leq |u_{n}|\leq \frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{(n-2)(n-1)n}$$=\frac{1}{2}\left[ \frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{(n-2)(n-1)}-\frac{1}{(n-1)n}{} \right]$$=\frac{1}{2}\left[ \frac{1}{2}-\frac{1}{(n-1)n}{} \right]$$=\frac{1}{4}-\frac{1}{2(n-1)n}<\frac{1}{4}\rightarrow u_{n}<\frac{1}{4}$.mong bạn đọc cho ý kiến!!!