\(\text { Tính giới hạn } A=\lim \left[\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+\ldots+\frac{1}{n(n+2)}\right] \text { . }\)
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Lời giải:
Báo sai\(\begin{aligned} &\text { Ta có } \frac{1}{n(n+2)}=\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+2}\right) . \\ &\text { Suy ra } \sum_{k=1}^{n} \frac{1}{k(k+2)}=\sum_{k=1}^{n} \frac{1}{2}\left(\frac{1}{k}-\frac{1}{k+2}\right)=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+\ldots+\frac{1}{n}-\frac{1}{n+2}\right) \\ &=\frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{n+1}-\frac{1}{n+2}\right) \\ &\text { Vậy } A=\lim \sum_{k=1}^{n} \frac{1}{k(k+2)}=\lim \frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{n+1}-\frac{1}{n+2}\right)=\lim \left(\frac{3}{4}-\frac{1}{2 n+2}-\frac{1}{2 n+4}\right)=\frac{3}{4} . \end{aligned}\)