Tìm \(L=\lim \left( \dfrac{1}{1}+\dfrac{1}{1+2}+\,...\,+\dfrac{1}{1+2+\,...\,+n} \right).\)

* Hướng dẫn giải

Ta có \(1+2+3+\,\,...\,\,+n=\frac{n\left( n+1 \right)}{2}\) (tổng của cấp số cộng với \({{u}_{1}}=1;\,\,d=1\)).

Khi đó \(\frac{1}{1+2}+\,...\,+\frac{1}{1+2+\,...\,+n}=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\,\,...\,\,+\frac{2}{n\left( n+1 \right)}=2\left( \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\,\,...\,\,+\frac{1}{n\left( n+1 \right)} \right)\)

\(=2\left( \frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+\,\,...\,\,+\frac{n+1-n}{n\left( n+1 \right)} \right)=2\left( \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\,\,...\,\,+\frac{1}{n}-\frac{1}{n+1} \right)=2\left( \frac{1}{2}-\frac{1}{n+1} \right).\)

Do đó \(\frac{1}{1}+\frac{1}{1+2}+\,...\,+\frac{1}{1+2+\,...\,+n}=1+2\left( \frac{1}{2}-\frac{1}{n+1} \right)=2-\frac{2}{n+1}=\frac{2n}{n+1}.\) Vậy \(L=\lim \frac{2n}{n+1}=2.\)

Chọn C