\u6570\u5217 \\(\\{ a _ n \\}\\) \u3092\r\n\\[\r\na _ 1 = 1 , \\ a _ {n+1} = \\sqrt{\\dfrac{3a _ n+4}{2a _ n+3}} \\quad ( n=1, 2, 3, \\cdots )\r\n\\]\r\n\u3067\u5b9a\u3081\u308b. \u4ee5\u4e0b\u306e\u554f\u3044\u306b\u7b54\u3048\u3088.<\/p>\r\n
(1)<\/strong>\u3000\\(n \\geqq 2\\) \u306e\u3068\u304d, \\(a _ n \\gt 1\\) \u3068\u306a\u308b\u3053\u3068\u3092\u793a\u305b.<\/p><\/li>\r\n (2)<\/strong>\u3000\\(\\alpha^2 = \\dfrac{3 \\alpha +4}{2 \\alpha +3}\\) \u3092\u6e80\u305f\u3059\u6b63\u306e\u5b9f\u6570 \\(\\alpha\\) \u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n (3)<\/strong>\u3000\u3059\u3079\u3066\u306e\u81ea\u7136\u6570 \\(n\\) \u306b\u5bfe\u3057\u3066, \\(a _ n \\lt \\alpha\\) \u3068\u306a\u308b\u3053\u3068\u3092\u793a\u305b.<\/p><\/li>\r\n (4)<\/strong>\u3000\\(0 \\lt r \\lt 1\\) \u3092\u6e80\u305f\u3059\u3042\u308b\u5b9f\u6570 \\(r\\) \u306b\u5bfe\u3057\u3066, \u4e0d\u7b49\u5f0f\r\n\\[\r\n\\dfrac{\\alpha -a _ {n+1}}{\\alpha -a _ n} \\leqq r \\quad ( n = 1, 2, 3, \\cdots )\r\n\\]\r\n\u304c\u6210\u308a\u7acb\u3064\u3053\u3068\u3092\u793a\u305b. \u3055\u3089\u306b, \u6975\u9650 \\(\\displaystyle\\lim _ {n \\rightarrow \\infty} a _ n\\) \u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n<\/ol>\r\n (1)<\/strong><\/p>\r\n \u6570\u5b66\u7684\u5e30\u7d0d\u6cd5\u3092\u7528\u3044\u3066\u793a\u3059.<\/p>\r\n 1*<\/strong>\u3000\\(n=2\\) \u306e\u3068\u304d\r\n\\[\r\na _ 2 = \\sqrt{\\dfrac{3+4}{2+3}} =\\sqrt{\\dfrac{7}{5}} \\gt 1\r\n\\]\r\n\u306a\u306e\u3067, \\(n=2\\) \u306e\u3068\u304d\u6210\u7acb\u3059\u308b.<\/p><\/li>\r\n 2*<\/strong>\u3000\\(n = k \\ ( k \\geqq 2 )\\) \u306e\u3068\u304d\u6210\u7acb\u3059\u308b, \u3059\u306a\u308f\u3061\r\n\\[\r\na _ k \\gt 1\\quad ... [1]\r\n\\]\r\n\u3068\u4eee\u5b9a\u3059\u308b\u3068\r\n\\[\\begin{align}\r\na _ {k+1} & = \\sqrt{\\dfrac{3}{2} -\\dfrac{1}{2( 2a _ k +3 )}} \\\\\r\n& \\gt \\sqrt{\\dfrac{3}{2} -\\dfrac{1}{2 \\cdot 5}} \\quad ( \\ \\text{\u2235} \\ [1] \\ ) \\\\\r\n& = \\sqrt{\\dfrac{7}{5}} \\gt 1 \\ .\r\n\\end{align}\\]\r\n\u3057\u305f\u304c\u3063\u3066, \\(n=k+1\\) \u306e\u3068\u304d\u3082\u6210\u7acb\u3059\u308b.<\/p><\/li>\r\n<\/ol>\r\n \u4ee5\u4e0a\u3088\u308a, \\(n \\geqq 2\\) \u306b\u5bfe\u3057\u3066\r\n\\[\r\na _ n \\gt 1\n\\]\r\n (2)<\/strong><\/p>\r\n \\[\\begin{align}\r\n\\alpha^2 & =\\dfrac{3 \\alpha +4}{2 \\alpha +3} \\\\\r\n2 \\alpha^3 +3 \\alpha^2 -3 \\alpha -4 & = 0 \\\\\r\n( \\alpha +1 )( 2 \\alpha^2 +\\alpha -4 ) & = 0 \\\\\r\n\\text{\u2234} \\quad \\alpha = -1 , & \\dfrac{-1 \\pm \\sqrt{33}}{4} \\ .\r\n\\end{align}\\]\r\n\\(\\alpha \\gt 0\\) \u306a\u306e\u3067\r\n\\[\r\n\\alpha = \\underline{\\dfrac{\\sqrt{33} -1}{4}}\n\\]\r\n (3)<\/strong><\/p>\r\n \u6570\u5b66\u7684\u5e30\u7d0d\u6cd5\u3092\u7528\u3044\u3066\u793a\u3059.<\/p>\r\n 1*<\/strong>\u3000\\(n=1\\) \u306e\u3068\u304d\r\n\\[\r\n\\alpha -a _ 1 = \\dfrac{\\sqrt{33} -5}{4} \\gt 0\n\\]\r\n\u306a\u306e\u3067, \\(n=1\\) \u306e\u3068\u304d\u306f\u6210\u7acb\u3059\u308b.<\/p><\/li>\r\n 2*<\/strong>\u3000\\(n = k \\ ( k \\geqq 1 )\\) \u306e\u3068\u304d\u6210\u7acb\u3059\u308b, \u3059\u306a\u308f\u3061\r\n\\[\r\n\\alpha -a _ k \\gt 0 \\quad ... [2]\r\n\\]\r\n\u3068\u4eee\u5b9a\u3059\u308b\u3068\r\n\\[\\begin{align}\r\n\\alpha^2 -{a _ {k+1}}^2 & = \\dfrac{3 \\alpha +4}{2 \\alpha +3} -\\dfrac{3a _ k+4}{2a _ k+3} \\\\\r\n& = \\dfrac{\\alpha -a _ k}{( 2 \\alpha +3 )( 2a _ k+3 )} \\\\\r\n& \\gt 0 \\quad ( \\ \\text{\u2235} \\ [2] \\ ) \\quad ... [3] \\ .\r\n\\end{align}\\]\r\n\u3057\u305f\u304c\u3063\u3066, \\(n=k+1\\) \u306e\u3068\u304d\u3082\u6210\u7acb\u3059\u308b.<\/p><\/li>\r\n<\/ol>\r\n \u4ee5\u4e0a\u3088\u308a, \u3059\u3079\u3066\u306e\u81ea\u7136\u6570 \\(n\\) \u306b\u5bfe\u3057\u3066\r\n\\[\r\na _ n \\lt \\alpha\n\\]\r\n (4)<\/strong><\/p>\r\n[3] \u306e\u9014\u4e2d\u7d4c\u904e\u3088\u308a\r\n\\[\\begin{align}\r\n\\alpha^2 -{a _ {k+1}}^2 & = \\dfrac{\\alpha -a _ k}{( 2 \\alpha +3 )( 2a _ k+3 )} \\\\\r\n\\text{\u2234} \\quad \\dfrac{\\alpha -a _ {k+1}}{\\alpha -a _ k} & = \\dfrac{1}{( \\alpha +a _ {k+1} )( 2 \\alpha +3 )( 2a _ k+3 )} \\quad ... [4] \\ .\r\n\\end{align}\\]\r\n[4] \u306e\u53f3\u8fba\u306e\u5206\u6bcd\u306b\u3064\u3044\u3066, \\(1 \\leqq a _ n \\lt \\alpha\\) \u306a\u306e\u3067\r\n\\[\\begin{align}\r\n( \\alpha & +a _ {k+1} )( 2 \\alpha +3 )( 2a _ k+3 ) \\\\\r\n& \\geqq ( 1+1 ) ( 2+3 ) ( 2+3 ) = 50 \\ .\r\n\\end{align}\\]\r\n\u3057\u305f\u304c\u3063\u3066, \\(r = \\dfrac{1}{50}\\) \u3068\u304a\u3051\u3070\r\n\\[\r\n\\dfrac{\\alpha -a _ {k+1}}{\\alpha -a _ k} \\leqq r\n\\]\r\n\u304c\u6210\u7acb\u3059\u308b.
\r\n\r\n\u3010 \u89e3 \u7b54 \u3011<\/h2>\r\n
\r\n
\r\n
\r\n\u3053\u3053\u304b\u3089\r\n\\[\r\n\\alpha -a _ {k+1} \\leqq r ( \\alpha -a _ k )\n\\]\r\n\u3053\u308c\u3092\u7e70\u8fd4\u3057\u7528\u3044\u308c\u3070, (3)<\/strong> \u306e\u7d50\u679c\u3068\u3042\u308f\u305b\u3066\r\n\\[\r\n0 \\lt \\alpha -a _ n \\leqq r ( \\alpha -a _ {n-1} ) \\leqq \\cdots \\leqq r^{n-1} ( \\alpha -a _ 1 )\n\\]\r\n\\(0 \\lt r \\lt 1\\) \u306a\u306e\u3067\r\n\\[\r\n\\displaystyle\\lim _ {n \\rightarrow \\infty} r^{n-1} ( \\alpha -a _ 1 ) = 0\n\\]\r\n\u3086\u3048\u306b, \u306f\u3055\u307f\u3046\u3061\u306e\u539f\u7406\u3088\u308a\r\n\\[\\begin{align}\r\n\\displaystyle\\lim _ {n \\rightarrow \\infty} ( \\alpha -a _ n ) & = 0 \\\\\r\n\\text{\u2234} \\quad \\displaystyle\\lim _ {n \\rightarrow \\infty} a _ n = \\alpha & = \\underline{\\dfrac{\\sqrt{33} -1}{4}} \\ .\r\n\\end{align}\\]\r\n","protected":false},"excerpt":{"rendered":"\u6570\u5217 \\(\\{ a _ n \\}\\) \u3092 \\[ a _ 1 = 1 , \\ a _ {n+1} = \\sqrt{\\dfrac{3a _ n+4}{2a _ n+3}} \\quad ( n=1, 2, 3, \\cdots … \u7d9a\u304d\u3092\u8aad\u3080