Với n là số nguyên dương, đặt...

Xét $A=\dfrac{1}{n\sqrt{n+1}+\left( n+1 \right)\sqrt{n}}$. Đặt $a=\sqrt{n},b=\sqrt{n+1}(a,b>0).~$
$A=\dfrac{1}{{{a}^{2}}b+{{b}^{2}}a}=\dfrac{1}{ab\left( a+b \right)}$
Ta có
${{b}^{2}}-{{a}^{2}}=n+1-n=1\Rightarrow \left( b-a \right)\left( b+a \right)=1\Rightarrow b-a=\dfrac{1}{a+b}$
Nên $A=\dfrac{b-a}{ab}=\dfrac{1}{a}-\dfrac{1}{b}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}$
Từ chứng mình trên ta có
${{S}_{n}}=\dfrac{1}{1\sqrt{2}+2\sqrt{1}}+\dfrac{1}{2\sqrt{3}+3\sqrt{2}}+...+\dfrac{1}{n\sqrt{n+1}+\left( n+1 \right)\sqrt{n}}$
$=\left( \dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}} \right)+\left( \dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}} \right)+...+\left( \dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}} \right)$
$=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}$
$=1-\dfrac{1}{\sqrt{n+1}}$
$\lim {{S}_{n}}=\lim \left( 1-\dfrac{1}{\sqrt{n+1}} \right)=\lim 1-\lim \left( \dfrac{1}{\sqrt{n+1}} \right)=1$