Với \(n\) là số tự nhiên lớn hơn 2, đặt \({S_n} = \frac{1}{{C_3^3}} + \frac{1}{{C_4^3}} + \frac{1}{{C_5^4}} + ... + \frac{1}{{C_n^3}}\).

* Hướng dẫn giải

Ta có: \(C_n^3 = \frac{{n!}}{{3!\left( {n - 3} \right)!}} = \frac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{6} \Rightarrow \frac{1}{{C_n^3}} = \frac{6}{{n\left( {n - 1} \right)\left( {n - 2} \right)}}\)

Khi đó:

\({S_n} = \frac{6}{{1.2.3}} + \frac{6}{{2.3.4}} + ... + \frac{6}{{\left( {n - 2} \right)\left( {n - 1} \right)n}} = 6\left( {\frac{1}{{1.2.3}} + \frac{1}{{2.3.4}} + ... + \frac{1}{{\left( {n - 2} \right)\left( {n - 1} \right)n}}} \right)\)

Xét dãy \((u_n)\): \({u_k} = \frac{1}{{\left( {k - 2} \right)\left( {k - 1} \right)k}} = \frac{1}{2}.\frac{1}{{k - 1}}\left( {\frac{1}{{k - 2}} - \frac{1}{k}} \right) = \frac{1}{2}\left( {\frac{1}{{k - 1}}.\frac{1}{{k - 2}} - \frac{1}{{k - 1}}.\frac{1}{k}} \right)\)

Suy ra

\(\begin{array}{l}
\frac{1}{{1.2.3}} = \frac{1}{2}\left( {\frac{1}{{1.2}} - \frac{1}{{2.3}}} \right) = \frac{1}{2}\left( {\frac{1}{2} - \frac{1}{6}} \right)\\
\frac{1}{{2.3.4}} = \frac{1}{2}\left( {\frac{1}{{2.4}} - \frac{1}{{3.4}}} \right) = \frac{1}{2}\left( {\frac{1}{6} - \frac{1}{{12}}} \right)\\
...\\
\frac{1}{{\left( {n - 2} \right)\left( {n - 1} \right)n}} = \frac{1}{2}\left( {\frac{1}{{n - 1}}.\frac{1}{{n - 2}} - \frac{1}{{n - 1}}.\frac{1}{n}} \right)
\end{array}\)

\( \Rightarrow {S_n} = 6.\frac{1}{2}\left( {\frac{1}{2} - \frac{1}{{n\left( {n - 1} \right)}}} \right) = 3\left( {\frac{1}{2} - \frac{1}{{n\left( {n - 1} \right)}}} \right)\)

Vậy \(\lim {S_n} = \lim \left[ {3\left( {\frac{1}{2} - \frac{1}{{n\left( {n - 1} \right)}}} \right)} \right] = \frac{3}{2}\)