Với n là số tự nhiên lớn hơn 2, $S_n$=$\frac{1}{3C3}+\frac{1}{4C3}+\frac{1}{5C4}+\cdots+\frac{1}{nC3}$ tính lim $S_n$

Ta có:
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
C_{n}^{3} =\frac{n!}{3!( n-3) !} =\frac{n( n-1)( n-2)}{6} \Rightarrow \frac{1}{C_{n}^{3}} =\frac{6}{n( n-1)( n-2)}\\
S_{n} =\frac{6}{1.2.3} +\frac{6}{2.3.4} +\frac{6}{3.4.5} +...+\frac{6}{( n-2)( n-1) n} =6\left(\frac{1}{1.2.3} +\frac{1}{2.3.4} +\frac{1}{3.4.5} +...+\frac{1}{( n-2)( n-1) n}\right)
\end{array}$
Xét dãy $\displaystyle ( u_{k}) :\ u_{k} =\frac{1}{( k-2)( k-1) k} =\frac{1}{2} .\frac{1}{k-1}\left(\frac{1}{k-2} -\frac{1}{k}\right) =\frac{1}{2}\left(\frac{1}{k-1} .\frac{1}{k-2} -\frac{1}{k-1} .\frac{1}{k}\right)$
Suy ra:
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
\frac{1}{1.2.3} =\frac{1}{2}\left(\frac{1}{1.2} -\frac{1}{2.3}\right) =\frac{1}{2}\left(\frac{1}{2} -\frac{1}{6}\right)\\
\frac{1}{2.3.4} =\frac{1}{2}\left(\frac{1}{2.3} -\frac{1}{3.4}\right) =\frac{1}{2}\left(\frac{1}{6} -\frac{1}{12}\right)\\
...\\
\frac{1}{( n-2)( n-1) n} =\frac{1}{2}\left(\frac{1}{( n-2)( n-1)} -\frac{1}{( n-1) n}\right)\\
\Rightarrow S_{n} =6.\frac{1}{2}\left(\frac{1}{2} -\frac{1}{n( n-1)}\right) =3\left(\frac{1}{2} -\frac{1}{n( n-1)}\right)
\end{array}$
Vậy $\displaystyle \lim S_{n} =\lim \left[ 3\left(\frac{1}{2} -\frac{1}{n( n-1)}\right)\right] =\frac{3}{2}$