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\(u_n=\dfrac{1}{\left(2-1\right)\left(2+1\right)}+\dfrac{1}{\left(3-1\right)\left(3+1\right)}+...+\dfrac{1}{\left(n-1\right)\left(n+1\right)}\)
\(u_n=\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+\dfrac{1}{4.6}+...+\dfrac{1}{\left(n-2\right)n}+\dfrac{1}{\left(n-1\right)\left(n+1\right)}\)
\(u_n=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{n-2}-\dfrac{1}{n}+\dfrac{1}{n-1}-\dfrac{1}{n+1}\right)\)
\(u_n=\dfrac{1}{2}\left(1+\dfrac{1}{2}-\dfrac{1}{n}-\dfrac{1}{n+1}\right)=\dfrac{1}{2}\left(\dfrac{3}{2}-\dfrac{1}{n}-\dfrac{1}{n+1}\right)\)
\(\Rightarrow lim\left(u_n\right)=lim\left(\dfrac{1}{2}\left(\dfrac{3}{2}-\dfrac{1}{n}-\dfrac{1}{n+1}\right)\right)=\dfrac{1}{2}.\dfrac{3}{2}=\dfrac{3}{4}\)
b/ \(u_n=\dfrac{1}{1^2+3}+\dfrac{1}{2^2+6}+...+\dfrac{1}{n^2+3n}=\dfrac{1}{1.4}+\dfrac{1}{2.5}+...+\dfrac{1}{n\left(n+3\right)}\)
\(u_n=\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{n}-\dfrac{1}{n+3}\right)\)
\(u_n=\dfrac{1}{3}\left(1+\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{n+1}-\dfrac{1}{n+2}-\dfrac{1}{n+3}\right)\)
\(\Rightarrow lim\left(u_n\right)=lim\left(\dfrac{1}{3}\left(1+\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{n+1}-\dfrac{1}{n+2}-\dfrac{1}{n+3}\right)\right)\)
\(\Rightarrow lim\left(u_n\right)=\dfrac{1}{3}\left(1+\dfrac{1}{2}+\dfrac{1}{3}\right)=\dfrac{11}{18}\)
\(\frac{n^3-1}{n^3+1}=\frac{\left(n-1\right)\left(n^2+n+1\right)}{\left(n+1\right)\left(n^2-n+1\right)}=\frac{\left(n-1\right)\left[\left(n+1\right)^2-\left(n+1\right)+1\right]}{\left(n+1\right)\left(n^2-n+1\right)}\)
\(\Rightarrow u_n=\frac{1.\left(3^2-3+1\right)}{3.\left(2^2-2+1\right)}.\frac{2\left(4^2-4+1\right)}{4.\left(3^2-3+1\right)}.\frac{3\left(5^2-5+1\right)}{5\left(4^2-4+1\right)}...\frac{\left(n-1\right)\left[\left(n+1\right)^2-\left(n+1\right)+1\right]}{\left(n+1\right)\left(n^2-n+1\right)}\)
\(\Rightarrow u_n=\frac{1.2.\left[\left(n+1\right)^2-\left(n+1\right)+1\right]}{\left(2^2-2+1\right).n\left(n+1\right)}=\frac{2n^2+2n+2}{3n^2+3n}\)
\(\Rightarrow lim\left(u_n\right)=lim\frac{2n^2+2n+2}{3n^2+3n}=\frac{2}{3}\)
\(\left\{{}\begin{matrix}u_1=a\\u_{n+1}=\frac{1}{2}u_n\end{matrix}\right.\)
\(\Rightarrow u_n\) là CSN với công bội \(q=\frac{1}{2}\)
\(\Rightarrow u_n=a.\left(\frac{1}{2}\right)^{n-1}\)
\(\Rightarrow lim\left(u_n\right)=lim\left(\frac{a}{2^{n-1}}\right)=0\)
Lời giải:
Xét hạng tử tổng quát:
\(\frac{1}{(n+1)\sqrt{n}+n\sqrt{n+1}}=\frac{(n+1)-n}{\sqrt{n(n+1)}(\sqrt{n}+\sqrt{n+1)}}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n(n+1)}}\)
\(=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Cho $n=1,2,...$ thì:
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}=1-\frac{1}{\sqrt{2}}\)
\(\frac{1}{3\sqrt{2}+2\sqrt{3}}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\)
......
\(\frac{1}{(n+1)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
\(\Rightarrow U_n=1-\frac{1}{\sqrt{n+1}}\)
\(\Rightarrow \lim\limits U_n=\lim (1-\frac{1}{\sqrt{n+1}})=1\)
Một câu thôi: Liên hợp
\(\dfrac{1}{2\sqrt{1}+\sqrt{2}}=\dfrac{2.1-\sqrt{2}}{2^2-2}=\dfrac{2-\sqrt{2}}{2}=1-\dfrac{1}{\sqrt{2}}\)
\(\dfrac{1}{3\sqrt{2}+2\sqrt{3}}=\dfrac{3\sqrt{2}-2\sqrt{3}}{9.2-4.3}=\dfrac{3\sqrt{2}-2\sqrt{3}}{6}=\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}\)
\(\Rightarrow\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
Nên chứng minh bằng quy nạp mạnh cho chặt chẽ, giờ tui buồn ngủ quá nên bạn tự chứng minh nha :(
\(\Rightarrow u_n=1-\dfrac{1}{\sqrt{n+1}}=\dfrac{\sqrt{n+1}-1}{\sqrt{n+1}}\Rightarrow\lim\limits\left(u_n\right)=\lim\limits\dfrac{\sqrt{\dfrac{n}{n}+\dfrac{1}{n}}-\dfrac{1}{\sqrt{n}}}{\sqrt{\dfrac{n}{n}+\dfrac{1}{n}}}=1\)
Sử dụng công thức tổng cấp số nhân:
\(1+3+3^2+...+3^n=\frac{3^{n+1}-1}{3-1}=\frac{3^{n+1}-1}{2}\)
\(1+4+...+4^n=\frac{4^{n+1}-1}{3}\)
\(\Rightarrow u_n=\frac{3\left(3^{n+1}-1\right)}{2\left(4^{n+1}-1\right)}=\frac{3.3^{n+1}-3}{2.4^{n+1}-2}\)
\(\Rightarrow lim\left(u_n\right)=lim\frac{3.3^{n+1}-3}{2.4^{n+1}-2}=\frac{3.\left(\frac{3}{4}\right)^{n+1}-3\left(\frac{1}{4}\right)^{n+1}}{2-2.\left(\frac{1}{4}\right)^{n+1}}=\frac{0}{2}=0\)
