Cho dãy số $\left( {{u}_{n}} \right)$ xác định bởi công thức...

Từ công thức xác định dãy $\left( {{u}_{n}} \right)$ suy ra ${{u}_{n}}>0$, $\forall n\in {{\mathbb{N}}^{*}}$
Ta có: ${{u}_{n+2}}=\dfrac{2{{u}_{n}}.{{u}_{n+1}}}{{{u}_{n}}+{{u}_{n+1}}}\Leftrightarrow \dfrac{1}{{{u}_{n+2}}}=\dfrac{1}{2}\left( \dfrac{1}{{{u}_{n+1}}}+\dfrac{1}{{{u}_{n}}} \right)$, $\forall n\in {{\mathbb{N}}^{*}}$
Đặt ${{v}_{n}}=\dfrac{1}{{{u}_{n}}}$, ta được ${{v}_{1}}=1$ ; ${{v}_{2}}=\dfrac{1}{2}$ và ${{v}_{n+2}}=\dfrac{1}{2}\left( {{v}_{n+1}}+{{v}_{n}} \right)$
$\Leftrightarrow {{v}_{n+2}}+\dfrac{1}{2}{{v}_{n+1}}={{v}_{n+1}}+\dfrac{1}{2}{{v}_{n}}$, $\forall n\in {{\mathbb{N}}^{*}}$
$\Rightarrow {{v}_{n+1}}+\dfrac{1}{2}{{v}_{n}}={{v}_{2}}+\dfrac{1}{2}{{v}_{1}}=1$, $\forall n\in {{\mathbb{N}}^{*}}$
$\Rightarrow {{v}_{n+1}}=-\dfrac{1}{2}{{v}_{n}}+1$, $\forall n\in {{\mathbb{N}}^{*}}$
$\Rightarrow {{v}_{n+1}}-\dfrac{2}{3}=-\dfrac{1}{2}\left( {{v}_{n}}-\dfrac{2}{3} \right)$, $\forall n\in {{\mathbb{N}}^{*}}$
Đặt ${{w}_{n}}={{v}_{n}}-\dfrac{2}{3}\Rightarrow \left( {{w}_{n}} \right)$ là một cấp số nhân với $\left\{ \begin{aligned}
& {{w}_{1}}=\dfrac{1}{3} \\
& q=-\dfrac{1}{2} \\
\end{aligned} \right.$
$\Rightarrow {{w}_{n}}=\dfrac{1}{3}.{{\left( -\dfrac{1}{2} \right)}^{n-1}}$
$\Rightarrow {{v}_{n}}=\dfrac{1}{3}{{\left( -\dfrac{1}{2} \right)}^{n-1}}+\dfrac{2}{3}$
$\Rightarrow {{u}_{n}}=\dfrac{1}{\dfrac{1}{3}{{\left( -\dfrac{1}{2} \right)}^{n-1}}+\dfrac{2}{3}}$
Vậy $\lim {{u}_{n}}=\lim \dfrac{1}{\dfrac{1}{3}{{\left( -\dfrac{1}{2} \right)}^{n-1}}+\dfrac{2}{3}}=\dfrac{3}{2}$.