\\(n\\) \u3092\u6b63\u306e\u6574\u6570\u3068\u3059\u308b. \u6570\u5217 \\(\\{ a _ k \\}\\) \u3092\r\n\\[\\begin{align}\r\na _ 1 & = \\dfrac{1}{n(n+1)} , \\\\\r\na _ {k+1} & = -\\dfrac{1}{k+n+1} +\\dfrac{n}{k} \\textstyle\\sum\\limits _ {i=1}^k a _ i \\quad ( k =1, 2, 3, \\cdots )\r\n\\end{align}\\]\r\n\u306b\u3088\u3063\u3066\u5b9a\u3081\u308b.<\/p>\r\n
(1)<\/strong>\u3000\\(a _ 2\\) \u304a\u3088\u3073 \\(a _ 3\\) \u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n (2)<\/strong>\u3000\u4e00\u822c\u9805 \\(a _ k\\) \u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n (3)<\/strong>\u3000\\(b _ n =\\textstyle\\sum\\limits _ {k=1}^n \\sqrt{a _ k}\\) \u3068\u304a\u304f\u3068\u304d, \\(\\displaystyle\\lim _ {n \\rightarrow \\infty} b _ n =\\log 2\\) \u3092\u793a\u305b.<\/p><\/li>\r\n<\/ol>\r\n (1)<\/strong><\/p>\r\n \\[\\begin{align}\r\na _ 2 & = -\\dfrac{1}{n+2} +\\dfrac{n}{1} \\cdot \\dfrac{1}{n(n+1)} \\\\\r\n& = \\underline{\\dfrac{1}{(n+1)(n+2)}} \\\\\r\na _ 3 & = -\\dfrac{1}{n+3} +\\dfrac{n}{2} \\left\\{ \\dfrac{1}{n(n+1)} +\\dfrac{1}{(n+1)(n+2)} \\right\\} \\\\\r\n& = -\\dfrac{1}{n+3} +\\dfrac{n}{2} \\cdot \\dfrac{2}{n(n+2)} \\\\\r\n& = \\underline{\\dfrac{1}{(n+2)(n+3)}}\r\n\\end{align}\\]\r\n (2)<\/strong><\/p>\r\n \u3059\u3079\u3066\u306e\u81ea\u7136\u6570 \\(k\\) \u306b\u3064\u3044\u3066\r\n\\[\r\na _ k = \\dfrac{1}{(n+k-1)(n+k)} \\quad ... [\\text{P}]\r\n\\]\r\n\u3067\u3042\u308b\u3053\u3068\u3092\u6570\u5b66\u7684\u5e30\u7d0d\u6cd5\u3092\u7528\u3044\u3066\u793a\u3059.<\/p>\r\n 1*<\/strong>\u3000\\(k=1\\) \u306e\u3068\u304d, \u6761\u4ef6\u3088\u308a [P] \u306f\u6210\u7acb\u3059\u308b.<\/p><\/li>\r\n 2*<\/strong>\u3000\\(1 \\leqq k \\leqq m \\ ( m \\geqq 1 )\\) \u306e\u3068\u304d, [P] \u304c\u6210\u7acb\u3059\u308b\u3068\u4eee\u5b9a\u3059\u308b\u3068\r\n\\[\\begin{align}\r\na _ {m+1} & = -\\dfrac{1}{n+m+1} \\\\\r\n& \\qquad +\\dfrac{n}{m} \\left\\{ \\dfrac{1}{n(n+1)} +\\cdots +\\dfrac{1}{(n+m-1)(n+m)} \\right\\} \\\\\r\n& = -\\dfrac{1}{n+m+1} +\\dfrac{n}{m} \\cdot \\dfrac{m}{n(n+m)} \\\\\r\n& = \\dfrac{1}{(n+m)(n+m+1)}\r\n\\end{align}\\]\r\n\u3057\u305f\u304c\u3063\u3066, \\(k = m+1\\) \u306e\u3068\u304d\u3082 [P] \u304c\u6210\u7acb\u3059\u308b.<\/p><\/li>\r\n<\/ol>\r\n 1*<\/strong> 2*<\/strong>\u3088\u308a, \u984c\u610f\u306f\u793a\u3055\u308c\u305f.<\/p>\r\n (3)<\/strong><\/p>\r\n (2)<\/strong> \u306e\u7d50\u679c\u3088\u308a\r\n\\[\r\n\\dfrac{1}{n+k} \\lt \\sqrt{a _ k} \\lt \\dfrac{1}{n+k-1}\r\n\\]\r\n\u3053\u308c\u306b \\(k= 1, 2 , \\cdots , n\\) \u3092\u4ee3\u5165\u3057\u305f\u5f0f\u3092\u8db3\u3057\u5408\u308f\u305b\u308b\u3068\r\n\\[\r\n\\underline{\\textstyle\\sum\\limits _ {k=1}^n \\dfrac{1}{n+k}} _ {[1]} \\lt b _ n \\lt \\underline{\\textstyle\\sum\\limits _ {k=1}^n \\dfrac{1}{n+k-1}} _ {[2]}\r\n\\]\r\n\u3053\u3053\u3067, [1] [2] \u306b\u3064\u3044\u3066, \\(n \\rightarrow \\infty\\) \u306e\u6975\u9650\u3092\u8003\u3048\u308b\u3068\r\n\\[\\begin{align}\r\n[1] & = \\dfrac{1}{n}\\displaystyle\\sum _ {k=1}^n \\dfrac{1}{1 +\\frac{k}{n}} \\\\\r\n& \\rightarrow \\displaystyle\\int _ 0^1 \\dfrac{dx}{1+x} =\\left[ \\log (1+x) \\right] _ 0^1 \\\\\r\n& = \\log 2 , \\\\\r\n[2] & = \\dfrac{1}{n} +\\dfrac{1}{n}\\displaystyle\\sum _ {k=1}^n \\dfrac{1}{1 +\\frac{k}{n}} -\\dfrac{1}{2n} \\\\\r\n& \\rightarrow 0 +\\displaystyle\\int _ 0^1 \\dfrac{dx}{1+x} -0 \\\\\r\n& = \\log 2\r\n\\end{align}\\]\r\n\u3088\u3063\u3066, \u306f\u3055\u307f\u3046\u3061\u306e\u539f\u7406\u3088\u308a\r\n\\[\r\n\\displaystyle\\lim _ {n \\rightarrow \\infty} b _ n = \\underline{\\log 2}\r\n\\]\r\n","protected":false},"excerpt":{"rendered":"\\(n\\) \u3092\u6b63\u306e\u6574\u6570\u3068\u3059\u308b. \u6570\u5217 \\(\\{ a _ k \\}\\) \u3092 \\[\\begin{align} a _ 1 & = \\dfrac{1}{n(n+1)} , \\\\ a _ {k+1} & = -\\dfrac{1} … \u7d9a\u304d\u3092\u8aad\u3080
\r\n\r\n\u3010 \u89e3 \u7b54 \u3011<\/h4>\r\n
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