Đáp án:
$\lim S_n = 1$
Giải thích các bước giải:
$\begin{array}{l}\text{Ta có:}\\ S_n = \dfrac{1}{1\sqrt2 + 2\sqrt1}+\dfrac{1}{2\sqrt3 + 3\sqrt2} + \cdots + \dfrac{1}{n\sqrt{n+1} + (n+1)\sqrt n}\\ \to S_n = \dfrac{1}{\sqrt{1.2}(\sqrt1 + \sqrt2)} + \dfrac{1}{\sqrt{2.3}(\sqrt2 + \sqrt3)} + \cdots + \dfrac{1}{\sqrt{n(n+1)}(\sqrt n + \sqrt{n+1})}\\ \to S_n = \dfrac{\sqrt2 - \sqrt 1}{\sqrt{1.2}} + \dfrac{\sqrt3 - \sqrt2}{\sqrt{2.3}} + \cdots + \dfrac{\sqrt{n+1} - \sqrt n}{\sqrt{n(n+1)}}\\ \to S_n = 1 - \dfrac{1}{\sqrt2} + \dfrac{1}{\sqrt2} - \dfrac{1}{\sqrt3} + \cdots + \dfrac{1}{\sqrt n} - \dfrac{1}{\sqrt{n+1}}\\ \to S_n = 1 - \dfrac{1}{\sqrt{n+1}}\\ \text{Khi đó:}\\ \quad \lim S_n = \lim\left(1 - \dfrac{1}{\sqrt{n+1}}\right)\\ \to \lim S_n = 1 -\lim\dfrac{1}{\sqrt{n+1}}\\ \to \lim S_n = 1 - 0 =1 \end{array}$
