(1)<\/strong>\u3000\u4e0d\u7b49\u5f0f \\(1-{q _ n}^2 \\leqq \\dfrac{(e-1)^2}{n^2}\\) \u3092\u793a\u3059\u3053\u3068\u306b\u3088\u308a, \\(\\displaystyle\\lim _ {n \\rightarrow \\infty} q _ n =1\\) \u3092\u8a3c\u660e\u305b\u3088. \u305f\u3060\u3057, \\(e\\) \u306f\u81ea\u7136\u5bfe\u6570\u306e\u5e95\u3067\u3042\u308b.<\/p><\/li>\r\n (2)<\/strong>\u3000\\(S _ n = \\displaystyle\\int _ {\\frac{1}{n}}^{p _ n} \\log (nx) \\, dx\\) \u3092 \\(p _ n\\) \u3067\u8868\u305b.<\/p><\/li>\r\n (3)<\/strong>\u3000\\(\\displaystyle\\lim _ {n \\rightarrow \\infty} n S _ n\\) \u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n<\/ol>\r\n (1)<\/strong><\/p>\r\n \\(C : \\ y=\\log (nx)\\) , \\(D : \\ \\left( x-\\dfrac{1}{n} \\right)^2+y^2=1\\) \u3068\u304a\u304f. (2)<\/strong><\/p>\r\n \\[\\begin{align}\r\nS _ n & = \\displaystyle\\int _ {\\frac{1}{n}}^{p _ n} \\log (nx) \\, dx \\\\\r\n& = \\big[ x \\log (nx) \\big] _ {\\frac{1}{n}}^{p _ n} -\\displaystyle\\int _ {\\frac{1}{n}}^{p _ n} x \\cdot \\dfrac{1}{x} \\, dx \\\\\r\n& = p _ n \\log \\left( n p _ n \\right) -\\big[ x \\big] _ {\\frac{1}{n}}^{p _ n} \\\\\r\n& = \\underline{p _ n \\left\\{ \\log \\left( n p _ n \\right) -1 \\right\\} +\\dfrac{1}{n}}\n\\end{align}\\]\r\n (3)<\/strong><\/p>\r\n \\(C\\) \u306e\u5f0f\u3088\u308a\r\n\\[\\begin{align}\r\nq _ n & = \\log \\left( n p _ n \\right) \\\\\r\n\\text{\u2234} \\quad n p _ n & = e^{q _ n}\n\\end{align}\\]\r\n(1)<\/strong> \u306e\u7d50\u679c\u3088\u308a, \\(\\displaystyle\\lim _ {n \\rightarrow \\infty} q _ n = 1\\) \u306a\u306e\u3067\r\n\\[\r\n\\displaystyle\\lim _ {n \\rightarrow \\infty} n p _ n = e\n\\]\r\n\u3053\u308c\u3092\u7528\u3044\u308c\u3070\r\n\\[\\begin{align}\r\nn S _ n & = n p _ n \\left\\{ \\log \\left( n p _ n \\right) -1 \\right\\} +1 \\\\\r\n& \\rightarrow e \\left( \\log e -1 \\right) +1 \\quad ( \\ \\text{\u2235} \\ n \\rightarrow \\infty \\text{\u306e\u3068\u304d} ) \\\\\r\n& = 1\n\\end{align}\\]\r\n\u3088\u3063\u3066\r\n\\[\r\n\\displaystyle\\lim _ {n \\rightarrow \\infty} n S _ n = \\underline{1}\n\\]\r\n","protected":false},"excerpt":{"rendered":"\\(n=1, 2, 3, \\cdots\\) \u306b\u5bfe\u3057\u3066, \\(y=\\log (nx)\\) \u3068 \\(\\left( x-\\dfrac{1}{n} \\right)^2+y^2=1\\) \u306e\u4ea4\u70b9\u306e\u3046\u3061\u7b2c \\(1\\) \u8c61\u9650\u306b\u3042\u308b\u70b9\u3092 … \u7d9a\u304d\u3092\u8aad\u3080
\r\n\r\n\u3010 \u89e3 \u7b54 \u3011<\/h4>\r\n
![\"\"](\"\/\/www.roundown.net\/nyushi\/wp-content\/uploads\/osaka_r_200905_01.png\" "\\"osaka_r_200905_01\\"")
\r\n
\r\n\\(C\\) \u306e\u5f0f\u3088\u308a\r\n\\[\\begin{align}\r\ne^y & = nx \\\\\r\n\\text{\u2234} \\quad x & = \\dfrac{e^y}{n}\n\\end{align}\\]\r\n\u3053\u308c\u3092 \\(D\\) \u306e\u5f0f\u306b\u4ee3\u5165\u3057\u3066\r\n\\[\\begin{align}\r\n\\left( \\dfrac{e^y-1}{n} \\right)^2 +y^2 & = 1 \\\\\r\n\\text{\u2234} \\quad 1-y^2 & = \\dfrac{\\left( e^y-1 \\right)^2}{n^2}\n\\end{align}\\]\r\n\u3053\u306e\u65b9\u7a0b\u5f0f\u306e\u89e3\u306f \\(y=q _ n\\) \u3067\u3042\u308a, \u307e\u305f \\(D\\) \u306e\u5f0f\u3088\u308a \\(q _ n \\leqq 1\\) \u306a\u306e\u3067\r\n\\[\\begin{align}\r\n1 -{q _ n}^2 & = \\dfrac{\\left( e^{q _ n}-1 \\right)^2}{n^2} \\leqq \\dfrac{\\left( e-1 \\right)^2}{n^2} \\\\\r\n\\text{\u2234} \\quad & \\underline{1-{q _ n}^2 \\leqq \\dfrac{\\left( e-1 \\right)^2}{n^2}}\n\\end{align}\\]\r\n\\(1 -{q _ n}^2 \\geqq 0\\) \u3067\u3042\u308a\r\n\\[\r\n\\displaystyle\\lim _ {n \\rightarrow \\infty} \\dfrac{(e-1)^2}{n^2} = 0\n\\]\r\n\u306a\u306e\u3067, \u306f\u3055\u307f\u3046\u3061\u306e\u539f\u7406\u3088\u308a\r\n\\[\r\n\\displaystyle\\lim _ {n \\rightarrow \\infty} \\left( 1-{q _ n}^2 \\right) =0\n\\]\r\n\\(q _ n \\gt 0\\) \u306a\u306e\u3067\r\n\\[\\begin{align}\r\n\\displaystyle\\lim _ {n \\rightarrow \\infty} \\left( 1-q _ n \\right) & = 0 \\\\\r\n\\text{\u2234} \\quad \\displaystyle\\lim _ {n \\rightarrow \\infty} q _ n & = \\underline{1}\n\\end{align}\\]\r\n