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Theo công thức tổng CSN:
\(1+\frac{2}{3}+...+\left(\frac{2}{3}\right)^n=\frac{1-\left(\frac{2}{3}\right)^{n+1}}{1-\frac{2}{3}}=3-3.\left(\frac{2}{3}\right)^{n+1}\)
\(1+\frac{1}{5}+...+\left(\frac{1}{5}\right)^n=\frac{1-\left(\frac{1}{5}\right)^{n+1}}{1-\frac{1}{5}}=\frac{5}{4}-\frac{5}{4}\left(\frac{1}{5}\right)^{n+1}\)
\(\Rightarrow lim\frac{3-3\left(\frac{2}{3}\right)^{n+1}}{\frac{5}{4}-\frac{5}{4}\left(\frac{1}{5}\right)^{n+1}}=\frac{3}{\frac{5}{4}}=\frac{12}{5}\)
Câu 2:
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{n(n+1)}=\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{(n+1)-n}{n(n+1)}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...\frac{1}{n}-\frac{1}{n+1}\)
\(=1-\frac{1}{n+1}\)
\(\Rightarrow \lim_{n\to \infty}(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{n(n+1)})=\lim_{n\to \infty}(1-\frac{1}{n+1})=1-\lim_{n\to \infty}\frac{1}{n+1}=1-0=1\)
a) \(\lim \frac{{5n + 1}}{{2n}} = \lim \frac{{5 + \frac{1}{n}}}{2} = \frac{{5 + 0}}{2} = \frac{5}{2}\)
b) \(\lim \frac{{6{n^2} + 8n + 1}}{{5{n^2} + 3}} = \lim \frac{{6 + \frac{8}{n} + \frac{1}{{{n^2}}}}}{{5 + \frac{3}{{{n^2}}}}} = \frac{{6 + 0 + 0}}{{5 + 0}} = \frac{6}{5}\)
c) \(\lim \frac{{\sqrt {{n^2} + 5n + 3} }}{{6n + 2}} = \lim \frac{{\sqrt {1 + \frac{5}{n} + \frac{3}{{{n^2}}}} }}{{6 + \frac{2}{n}}} = \frac{{\sqrt {1 + 0 + 0} }}{{6 + 0}} = \frac{1}{6}\)
d) \(\lim \left( {2 - \frac{1}{{{3^n}}}} \right) = \lim 2 - \lim {\left( {\frac{1}{3}} \right)^n} = 2 - 0 = 0\)
e) \(\lim \frac{{{3^n} + {2^n}}}{{{{4.3}^n}}} = \lim \frac{{1 + {{\left( {\frac{2}{3}} \right)}^n}}}{4} = \frac{{1 + 0}}{4} = \frac{1}{4}\)
g) \(\lim \frac{{2 + \frac{1}{n}}}{{{3^n}}}\)
Ta có \(\lim \left( {2 + \frac{1}{n}} \right) = \lim 2 + \lim \frac{1}{n} = 2 + 0 = 2 > 0;\lim {3^n} = + \infty \Rightarrow \lim \frac{{2 + \frac{1}{n}}}{{{3^n}}} = 0\)
Bài 1. Ta có:
\(\begin{array}{l} S = \sum\limits_{k = 1}^n {{x^{2k}}} + \sum\limits_{k = 1}^n {\dfrac{1}{{{x^{2k}}}} + 2n} \\ = {x^2}\dfrac{{1 - {x^{2n}}}}{{1 - {x^2}}} + \dfrac{1}{{{x^2}}}.\dfrac{{1 - \dfrac{1}{{{x^{2n}}}}}}{{1 - \dfrac{1}{{{x^2}}}}} + 2n\\ = \dfrac{{\left( {1 - {x^{2n}}} \right)\left( {{x^{2n + 2}} - 1} \right)}}{{\left( {1 - {x^2}} \right){x^{2n}}}} + 2n \end{array}\)
Bài 2.
Ta có:
\(\begin{array}{l} T = \dfrac{1}{2} + \dfrac{3}{{{2^2}}} + \dfrac{5}{{{2^3}}} + ... + \dfrac{{2n - 1}}{{{2^n}}}\left( 1 \right)\\ \dfrac{1}{2}T = \dfrac{1}{{{2^2}}} + \dfrac{3}{{{2^3}}} + \dfrac{5}{{{2^4}}} + ... + \dfrac{{2n - 3}}{{{2^n}}} + \dfrac{{2n - 1}}{{{2^{n + 1}}}}\left( 2 \right) \end{array}\)
\((1)-(2)\)\(\Rightarrow \dfrac{1}{2}T = \dfrac{1}{2} + \dfrac{2}{{{2^2}}} + \dfrac{2}{{{2^3}}} + ... + \dfrac{2}{{{2^n}}} - \dfrac{{2n - 1}}{{{2^{n + 1}}}}\)
\(\begin{array}{l} \Rightarrow T = 2\left[ {\dfrac{1}{2} + \dfrac{1}{2}\dfrac{{1 - {{\left( {\dfrac{1}{2}} \right)}^{n - 1}}}}{{1 - \dfrac{1}{2}}} - \dfrac{{2n - 1}}{{{2^{n + 1}}}}} \right]\\ = 1 + \dfrac{{{2^{n - 1}} - 1}}{{{2^{n - 2}}}} - \dfrac{{2n - 1}}{{{2^n}}} \end{array}\)
\(S=x^2+\frac{1}{x^2}+2+x^4+\frac{1}{x^4}+2+...+x^{2n}+\frac{1}{x^{2n}}+2\)
\(=\left(x^2+x^4+...+x^{2n}\right)+\left(\frac{1}{x^2}+\frac{1}{x^4}+...+\frac{1}{x^{2n}}\right)+2n\)
\(=x^2.\frac{\left(x^2\right)^{n-1}-1}{x^2-1}+\frac{1}{x^2}.\frac{\left(\frac{1}{x^2}\right)^{n-1}-1}{\frac{1}{x^2}-1}+2n\)
\(=\frac{x^{2n}-x^2}{x^2-1}+\frac{x^{2-2n}-1}{1-x^2}+2n\)
\(T=\frac{1}{2}+\frac{3}{2^2}+\frac{5}{2^3}+...+\frac{2n-3}{2^{n-1}}+\frac{2n-1}{2^n}\)
\(\Rightarrow2T=1+\frac{3}{2}+\frac{5}{2^2}+...+\frac{2n-1}{2^{n-1}}\)
\(\Rightarrow T=1+\frac{2}{2}+\frac{2}{2^2}+\frac{2}{2^3}+...+\frac{2}{2^{n-1}}-\frac{2n-1}{2^n}\)
\(T=1+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{n-2}}-\frac{2n-1}{2^n}\)
\(T=1+1.\frac{\left(\frac{1}{2}\right)^{n-2}-1}{\frac{1}{2}-1}-\frac{2n-1}{2^n}=3-\frac{1}{2^{n-1}}-\frac{2n-1}{2^n}=3-\frac{1}{2^n}-\frac{n}{2^{n-1}}\)
Ta có:
\(u_1=\dfrac{1}{3^1-1}=\dfrac{1}{2}\\ u_2=\dfrac{2}{3^2-1}=\dfrac{1}{4}\\ u_3=\dfrac{3}{3^3-1}=\dfrac{3}{26}\)
\(\Rightarrow B\)
a) \(\lim\limits3=3\) vì \(3\) là hằng số.
Áp dụng giới hạn cơ bản với \(k=2\), ta có:\(\lim\limits\dfrac{1}{n^2}=0\).
b) \(\lim\limits\left(3+\dfrac{1}{n^2}\right)=\lim\limits3+\lim\limits\dfrac{1}{n^2}=3\).
Hỏng biết lun