Cho dãy số \(({u_n})\)thỏa mãn: \(\left\{ \begin{array}{l}{u_1} = 1\\{u_{n + 1}} = \sqrt {\frac{2}{3}{u_n}^2 + \frac{{n – 2}}{{{n^2} + n}}} \end{array} \right.;\forall n \in N*\)
Tìm công thức số hạng tổng quát của dãy số \(({u_n})\)và tính \(\lim {u_n}\).
Lời giải
+) Xét: \(u_{n + 1}^2 = \frac{2}{3}{u_n}^2 + \frac{{n – 2}}{{n(n + 1)}}\)\( = \frac{2}{3}.\left( {\frac{2}{3}u_{n – 1}^2 + \frac{{n – 1 – 2}}{{(n – 1)n}}} \right) + \frac{{n – 2}}{{n(n + 1)}}\)\( = {\left( {\frac{2}{3}} \right)^2}u_{n – 1}^2 + \frac{{n – 2}}{{n(n + 1)}} + \frac{2}{3}\frac{{n – 3}}{{(n – 1)n}}\)\( = {\left( {\frac{2}{3}} \right)^2}.\left( {\frac{2}{3}u_{n – 2}^2 + \frac{{n – 2 – 2}}{{(n – 2)(n – 1)}}} \right) + \frac{{n – 2}}{{n(n + 1)}} + \frac{2}{3}\frac{{n – 3}}{{(n – 1)n}}\)
\( = {\left( {\frac{2}{3}} \right)^3}.u_{n – 2}^2 + \frac{{n – 2}}{{n(n + 1)}} + \frac{2}{3}\frac{{n – 3}}{{(n – 1)n}} + {\left( {\frac{2}{3}} \right)^2}\frac{{n – 4}}{{(n – 2)(n – 1)}}\)\( = …\)
\( = {\left( {\frac{2}{3}} \right)^n}.{u_1} + \frac{{n – 2}}{{n(n + 1)}} + \frac{2}{3}.\frac{{n – 3}}{{(n – 1)n}} + {\left( {\frac{2}{3}} \right)^2}.\frac{{n – 4}}{{(n – 2)(n – 1)}} + … + {\left( {\frac{2}{3}} \right)^{n – 1}}.\frac{{ – 1}}{{1.2}}\)
\( \Rightarrow u_{n + 1}^2 = {\left( {\frac{2}{3}} \right)^n} + {\sum\limits_{i = 0}^{n – 1} {\left( {\frac{2}{3}} \right)} ^i}.\frac{{n – 2 – i}}{{(n – i)(n + 1 – i)}}\) =\({\left( {\frac{2}{3}} \right)^n} + S\) (1)
+) Xét hai số hạng liền kề của tổng \(S = {\sum\limits_{i = 0}^{n – 1} {\left( {\frac{2}{3}} \right)} ^i}.\frac{{n – 2 – i}}{{(n – i)(n + 1 – i)}}\):
\({S_k} = {\left( {\frac{2}{3}} \right)^k}.\frac{{n – 2 – k}}{{(n – k)(n + 1 – k)}} + {\left( {\frac{2}{3}} \right)^{k + 1}}.\frac{{n – 3 – k}}{{(n – k – 1)(n – k)}}\) \(\left( {k = 0;1;…;n – 2} \right)\)
\( = {\left( {\frac{2}{3}} \right)^k}.\left( {\frac{{n – 2 – k}}{{(n – k)(n + 1 – k)}} + \frac{2}{3}.\frac{{n – 3 – k}}{{(n – k – 1)(n – k)}}} \right)\)
\( = {\left( {\frac{2}{3}} \right)^k}.\left( {\frac{{n – 2 – k}}{{n – k}} – \frac{{n – 2 – k}}{{n + 1 – k}} + \frac{2}{3}.\frac{{n – 3 – k}}{{n – k – 1}} – \frac{2}{3}.\frac{{n – 3 – k}}{{n – k}}} \right)\)
\( = {\left( {\frac{2}{3}} \right)^k}.\left( {\left( {\frac{{n – 2 – k}}{{n – k}} – \frac{2}{3}.\frac{{n – 3 – k}}{{n – k}}} \right) – \frac{{n – 2 – k}}{{n + 1 – k}} + \frac{2}{3}.\frac{{n – 3 – k}}{{n – k – 1}}} \right)\)
\( = {\left( {\frac{2}{3}} \right)^k}.\left( {\frac{1}{3} – \frac{{n – 2 – k}}{{n + 1 – k}} + \frac{2}{3}.\frac{{n – 3 – k}}{{n – k – 1}}} \right)\)
\( = {\left( {\frac{2}{3}} \right)^k}.\left( {\frac{1}{3} – 1 + \frac{3}{{n + 1 – k}} + \frac{2}{3} – \frac{2}{3}.\frac{2}{{n – k – 1}}} \right)\)
\({S_k} = {\left( {\frac{2}{3}} \right)^k}.\left( {\frac{3}{{n + 1 – k}} – \frac{4}{{3(n – k – 1)}}} \right)\)
+) Ta thấy số \({S_k}\) gồm hai hạng tử, hạng tử đầu tiên của \({S_{k + 2}}\) và hạng tử thứ hai của \({S_k}\) có tổng bằng 0, vì:
\({\left( {\frac{2}{3}} \right)^{k + 2}}.\frac{3}{{n – k – 1}} – {\left( {\frac{2}{3}} \right)^k}.\frac{4}{{3(n – k – 1)}} = 0\)
\( \Rightarrow \sum\limits_{k = 0}^{n – 2} {{s_k}} = {\left( {\frac{2}{3}} \right)^0}.\frac{3}{{n + 1}} + {\left( {\frac{2}{3}} \right)^1}.\frac{3}{n} – {\left( {\frac{2}{3}} \right)^{n – 3}}.\frac{4}{{3.2}} – {\left( {\frac{2}{3}} \right)^{n – 2}}.\frac{4}{{3.1}}\) (2)
+) Từ đẳng thức bổ đề: \(S = {A_0} + {A_1} + … + {A_{n – 1}} \Rightarrow 2S = {A_0} + {A_{n – 1}} + \sum\limits_{k = 0}^{n – 2} {({A_k} + {A_{k + 1}})} \) (3)
(1), (3) \( \Rightarrow 2u_{n + 1}^2 = 2.{\left( {\frac{2}{3}} \right)^n} + 2S = 2.{\left( {\frac{2}{3}} \right)^n} + \frac{{n – 2}}{{n(n + 1)}} + {\left( {\frac{2}{3}} \right)^{n – 1}}.\frac{{ – 1}}{2} + \sum\limits_{k = 0}^{n – 2} {{s_k}} \)
(2) \( \Rightarrow 2u_{n + 1}^2 = 2.{\left( {\frac{2}{3}} \right)^n} + \frac{{n – 2}}{{n(n + 1)}} + {\left( {\frac{2}{3}} \right)^{n – 1}}.\frac{{ – 1}}{2} + {\left( {\frac{2}{3}} \right)^0}.\frac{3}{{n + 1}} + {\left( {\frac{2}{3}} \right)^1}.\frac{3}{n} – {\left( {\frac{2}{3}} \right)^{n – 3}}.\frac{4}{{3.2}} – {\left( {\frac{2}{3}} \right)^{n – 2}}.\frac{4}{{3.1}}\)
\( \Leftrightarrow 2u_{n + 1}^2 = \left( {2.{{\left( {\frac{2}{3}} \right)}^n} – \frac{1}{2}.{{\left( {\frac{2}{3}} \right)}^{n – 1}} – {{\left( {\frac{2}{3}} \right)}^{n – 2}} – \frac{4}{3}.{{\left( {\frac{2}{3}} \right)}^{n – 2}}} \right) + \left( {\frac{4}{{3.1}} + \frac{{n – 2}}{{n(n + 1)}} + \frac{3}{{n + 1}} + \frac{2}{n}} \right)\)
\( \Leftrightarrow 2u_{n + 1}^2 = – 4.{\left( {\frac{2}{3}} \right)^n} + \frac{6}{{n + 1}}\)\( \Leftrightarrow u_{n + 1}^2 = \frac{3}{{n + 1}} – 2.{\left( {\frac{2}{3}} \right)^n}\)
\( \Leftrightarrow {u_{n + 1}} = \sqrt {\frac{3}{{n + 1}} – 2.{{\left( {\frac{2}{3}} \right)}^n}} \).
Vậy công thức tổng quát của dãy là\({u_{n + 1}} = \sqrt {\frac{3}{{n + 1}} – 2.{{\left( {\frac{2}{3}} \right)}^n}} \).
+) \(u_{n + 1}^2 = \frac{2}{3}{u_n}^2 + \frac{{n – 2}}{{n(n + 1)}} \ge 0\) \((\forall n \ge 2)\) ; \({u_1} = 1 \ge 0\) (4)
Xét \(f(x) = \frac{3}{{x + 1}} – 2.{\left( {\frac{2}{3}} \right)^x}\) \( \Rightarrow f'(x) = \frac{{ – 3}}{{{{\left( {x + 1} \right)}^2}}} – 2.{\left( {\frac{2}{3}} \right)^x}.\ln \frac{2}{3} < 0\) (5)
+) Từ (4) và (5) \( \Rightarrow \)dãy \({u_{n + 1}}\) là dãy dương giảm.
\( \Rightarrow {\lim _{n \to + \infty }}{u_{n + 1}} = \sqrt {\frac{3}{{n + 1}} – 2.{{\left( {\frac{2}{3}} \right)}^n}} = \sqrt {0 – 0} = 0\).
Trả lời