Đáp án:
23. $\dfrac34$
24. $\dfrac{11}{8}$
Giải thích các bước giải:
$\begin{array}{l}23.\quad \lim\left[\dfrac{1}{1.3} + \dfrac{1}{2.4} + \cdots + \dfrac{1}{n(n+2)} \right]\\ =\lim\left\{\dfrac{1}{2}\left[\dfrac{2}{1.3} + \dfrac{2}{2.4} + \cdots + \dfrac{2}{n(n+2)} \right]\right\}\\ = \lim\left[\dfrac12\left(1 - \dfrac13 + \dfrac12 - \dfrac14 + \cdots + \dfrac1n - \dfrac{1}{n+2} \right) \right]\\ = \lim\left[\dfrac12\left( 1 + \dfrac12 - \dfrac{1}{n+1} -\dfrac{1}{n+2}\right)\right]\\ = \dfrac{3}{4} - \lim\dfrac{2n+3}{2(n+1)(n+2)}\\ = \dfrac{3}{4} - \lim\dfrac{\dfrac2n+\dfrac{3}{n^2}}{2\left(1+\dfrac1n\right)\left(1 + \dfrac2n\right)}\\ = \dfrac34 - \dfrac{2.0 + 3.0}{2(1.0)(1 + 2.0)}\\ = \dfrac34\\ 24.\quad \lim\left[\dfrac{1}{1.4} + \dfrac{1}{2.5} + \cdots + \dfrac{1}{n(n+3)} \right]\\ =\lim\left\{\dfrac{1}{3}\left[\dfrac{3}{1.4} + \dfrac{3}{2.5} + \cdots + \dfrac{3}{n(n+3)} \right]\right\}\\ = \lim\left[\dfrac13\left(1 - \dfrac14 + \dfrac12 - \dfrac15 + \cdots + \dfrac1n - \dfrac{1}{n+3} \right) \right]\\ = \lim\left[\dfrac13\left(1 + \dfrac12+\dfrac13 -\dfrac{1}{n+1} -\dfrac{1}{n+2}- \dfrac{1}{n+3}\right)\right]\\ = \dfrac{11}{8} - \lim\dfrac{3n^2 + 12n +11}{3(n+1)(n+2)(n+3)}\\ = \dfrac{11}{8} - \lim\dfrac{\dfrac3n +\dfrac{12}{n}+\dfrac{11}{n^3}}{3\left(1+\dfrac1n\right)\left(1+\dfrac2n\right)\left(1 + \dfrac3n\right)}\\ = \dfrac{11}{8} - \dfrac{3.0 + 12.0 + 11.0}{3(1+0)(1+2.0)(1 + 3.0)}\\ = \dfrac{11}{8}\\ \end{array}$
