Toán Tìm L= lim (1/1+1/1+2+…..+1/1+2+…..+n) Giúp mình với ạ. Mình cảm ơn 05/10/2021 By Arya Tìm L= lim (1/1+1/1+2+…..+1/1+2+…..+n) Giúp mình với ạ. Mình cảm ơn
Đáp án: \(L = 2\) Giải thích các bước giải: Ta có: \(\begin{array}{l} 1 + 2 + 3 + …. + n = \dfrac{{n\left( {n + 1} \right)}}{2}\\ \dfrac{1}{{n\left( {n + 1} \right)}} = \dfrac{{\left( {n + 1} \right) – n}}{{n\left( {n + 1} \right)}} = \dfrac{1}{n} – \dfrac{1}{{n + 1}}\\ L = \lim \left( {\dfrac{1}{1} + \dfrac{1}{{1 + 2}} + \dfrac{1}{{1 + 2 + 3}} + …… + \dfrac{1}{{1 + 2 + 3 + …. + n}}} \right)\\ = \lim \left[ {\dfrac{1}{1} + \dfrac{1}{{\dfrac{{2.3}}{2}}} + \dfrac{1}{{\dfrac{{3.4}}{2}}} + …… + \dfrac{1}{{\dfrac{{n\left( {n + 1} \right)}}{2}}}} \right]\\ = \lim \left[ {\dfrac{1}{1} + \dfrac{2}{{2.3}} + \dfrac{2}{{3.4}} + …… + \dfrac{2}{{n\left( {n + 1} \right)}}} \right]\\ = \lim \left[ {1 + 2.\left( {\dfrac{1}{{2.3}} + \dfrac{1}{{3.4}} + ….. + \dfrac{1}{{n\left( {n + 1} \right)}}} \right)} \right]\\ = \lim \left[ {1 + 2.\left( {\dfrac{1}{2} – \dfrac{1}{3} + \dfrac{1}{3} – \dfrac{1}{4} + ….. + \dfrac{1}{n} – \dfrac{1}{{n + 1}}} \right)} \right]\\ = \lim \left[ {1 + 2.\left( {\dfrac{1}{2} – \dfrac{1}{{n + 1}}} \right)} \right]\\ = \lim \left( {2 – \dfrac{2}{{n + 1}}} \right)\\ = 2 – 0 = 2 \end{array}\) Trả lời
Đáp án: \(L = 2\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l} 1 + 2 + 3 + …. + n = \dfrac{{n\left( {n + 1} \right)}}{2}\\ \dfrac{1}{{n\left( {n + 1} \right)}} = \dfrac{{\left( {n + 1} \right) – n}}{{n\left( {n + 1} \right)}} = \dfrac{1}{n} – \dfrac{1}{{n + 1}}\\ L = \lim \left( {\dfrac{1}{1} + \dfrac{1}{{1 + 2}} + \dfrac{1}{{1 + 2 + 3}} + …… + \dfrac{1}{{1 + 2 + 3 + …. + n}}} \right)\\ = \lim \left[ {\dfrac{1}{1} + \dfrac{1}{{\dfrac{{2.3}}{2}}} + \dfrac{1}{{\dfrac{{3.4}}{2}}} + …… + \dfrac{1}{{\dfrac{{n\left( {n + 1} \right)}}{2}}}} \right]\\ = \lim \left[ {\dfrac{1}{1} + \dfrac{2}{{2.3}} + \dfrac{2}{{3.4}} + …… + \dfrac{2}{{n\left( {n + 1} \right)}}} \right]\\ = \lim \left[ {1 + 2.\left( {\dfrac{1}{{2.3}} + \dfrac{1}{{3.4}} + ….. + \dfrac{1}{{n\left( {n + 1} \right)}}} \right)} \right]\\ = \lim \left[ {1 + 2.\left( {\dfrac{1}{2} – \dfrac{1}{3} + \dfrac{1}{3} – \dfrac{1}{4} + ….. + \dfrac{1}{n} – \dfrac{1}{{n + 1}}} \right)} \right]\\ = \lim \left[ {1 + 2.\left( {\dfrac{1}{2} – \dfrac{1}{{n + 1}}} \right)} \right]\\ = \lim \left( {2 – \dfrac{2}{{n + 1}}} \right)\\ = 2 – 0 = 2 \end{array}\)