Tính \({\rm{lim}}\left( {\frac{1}{{{n^2} + 1}} + \frac{2}{{{n^2} + 2}} + \ldots + \frac{n}{{{n^2} + n}}} \right)\) bằng

Giải thích

Ta có: \(\frac{{{n^2} + n}} \le \frac{1}{{{n^2} + 1}} + \frac{2}{{{n^2} + 2}} +  \ldots  + \frac{n}{{{n^2} + n}} \le \frac{{{n^2} + 1}}\).

Mà \({\rm{lim}}\frac{{{n^2} + n}} = {\rm{lim}}\frac{{\frac{{n\left( {n + 1} \right)}}{2}}}{{{n^2} + n}} = \frac{1}{2}\);

\({\rm{lim}}\frac{{{n^2} + 1}} = {\rm{lim}}\frac{{\frac{{n\left( {n + 1} \right)}}{2}}}{{{n^2} + 1}} = {\rm{lim}}\frac{{{n^2} + n}}{{2\left( {{n^2} + 1} \right)}} = {\rm{lim}}\frac}{{2\left( {1 + \frac{1}{{{n^2}}}} \right)}} = \frac{1}{2}\)

Nên \({\rm{lim}}\left( {\frac{1}{{{n^2} + 1}} + \frac{2}{{{n^2} + 2}} +  \ldots  + \frac{n}{{{n^2} + n}}} \right) = \frac{1}{2}\).

 Chọn C