\u81ea\u7136\u6570 \\(n\\) \u306b\u5bfe\u3057\r\n\\[\\begin{align}\r\nS _ n & = \\displaystyle\\int _ 0^1 \\dfrac{1 -(-x)^n}{1+x} \\, dx \\\\\r\nT _ n & = \\textstyle\\sum\\limits _ {k=1}^n \\dfrac{(-1)^{k-1}}{k( k+1 )}\r\n\\end{align}\\]\r\n\u3068\u304a\u304f. \u3053\u306e\u3068\u304d\u4ee5\u4e0b\u306e\u5404\u554f\u3044\u306b\u7b54\u3048\u3088.<\/p>\r\n
(2)<\/strong>\u3000\\(T _ n -2S _ n\\) \u3092 \\(n\\) \u3092\u7528\u3044\u3066\u8868\u305b.<\/p><\/li>\r\n (3)<\/strong>\u3000\u6975\u9650\u5024 \\(\\displaystyle\\lim _ {n \\rightarrow \\infty} T _ n\\) \u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n<\/ol>\r\n (1)<\/strong><\/p>\r\n \\(0 \\leqq x \\leqq 1\\) \u306b\u304a\u3044\u3066, \\(\\dfrac{x^n}{1+x} \\leqq x^n\\) \u3067\u3042\u308b\u3053\u3068\u3092\u7528\u3044\u3066\r\n\\[\\begin{align}\r\n\\left| S _ n -\\displaystyle\\int _ 0^1 \\dfrac{1}{1+x} \\, dx \\right| & = \\left| \\displaystyle\\int _ 0^1 \\dfrac{(-x)^n}{1+x} \\, dx \\right| \\\\\r\n& = \\displaystyle\\int _ 0^1 \\dfrac{n^n}{1+x} \\, dx \\leqq \\displaystyle\\int _ 0^1 x^n \\, dx \\\\\r\n& = \\left[ \\dfrac{x^{n+1}}{n+1} \\right] _ 0^1 = \\dfrac{1}{n+1}\n\\end{align}\\]\r\n\u3086\u3048\u306b\r\n\\[\r\n\\left| S _ n -\\displaystyle\\int _ 0^1 \\dfrac{1}{1+x} \\, dx \\right| \\leqq \\dfrac{1}{n+1}\n\\]\r\n (2)<\/strong><\/p>\r\n \\[\\begin{align}\r\nS _ n & = \\displaystyle\\int _ 0^1 \\dfrac{1-(-x)^n}{1-(-x)} \\, dx = \\textstyle\\sum\\limits _ {k=1}^n \\displaystyle\\int _ 0^1 (-x)^{k-1} \\, dx \\\\\r\n& = \\textstyle\\sum\\limits _ {k=1}^n \\left[ -\\dfrac{(-x)^k}{k} \\right] _ 0^1 \\\\\r\n& = \\textstyle\\sum\\limits _ {k=1}^n \\dfrac{(-1)^{k-1}}{k}\n\\end{align}\\]\r\n\u307e\u305f\r\n\\[\\begin{align}\r\nT _ n & = \\textstyle\\sum\\limits _ {k=1}^n (-1)^{k-1} \\left( \\dfrac{1}{k} -\\dfrac{1}{k+1} \\right) \\\\\r\n& = \\textstyle\\sum\\limits _ {k=1}^n \\dfrac{(-1)^{k-1}}{k} +\\textstyle\\sum\\limits _ {k=2}^{n+1} \\dfrac{(-1)^{k-1}}{k}\n\\end{align}\\]\r\n\u3053\u308c\u3089\u3092\u7528\u3044\u308c\u3070\r\n\\[\\begin{align}\r\nT _ n -2S _ n & = \\textstyle\\sum\\limits _ {k=2}^{n+1} \\dfrac{(-1)^{k-1}}{k} -\\textstyle\\sum\\limits _ {k=1}^n \\dfrac{(-1)^{k-1}}{k} \\\\\r\n& = \\underline{\\dfrac{(-1)^n}{n+1}-1}\n\\end{align}\\]\r\n (3)<\/strong><\/p>\r\n (1)<\/strong> \u306e\u7d50\u679c\u306b\u3064\u3044\u3066, \\(\\displaystyle\\lim _ {n \\rightarrow \\infty} \\dfrac{1}{n+1} =0\\) \u306a\u306e\u3067, \u306f\u3055\u307f\u3046\u3061\u306e\u539f\u7406\u3088\u308a\r\n\\[\\begin{align}\r\n\\displaystyle\\lim _ {n \\rightarrow \\infty} & \\left| S _ n -\\displaystyle\\int _ 0^1 \\dfrac{1}{1+x} \\, dx \\right| = 0 \\\\\r\n\\text{\u2234} \\quad \\displaystyle\\lim _ {n \\rightarrow \\infty} S _ n & = \\displaystyle\\int _ 0^1 \\dfrac{1}{1+x} \\, dx \\\\\r\n& = \\big[ \\log (1+x) \\big] _ 0^1 = \\log 2\n\\end{align}\\]\r\n (2)<\/strong> \u3088\u308a\r\n\\[\\begin{align}\r\n\\displaystyle\\lim _ {n \\rightarrow \\infty} T _ n & = \\displaystyle\\lim _ {n \\rightarrow \\infty} \\left\\{ 2S _ n +\\dfrac{(-1)^n}{n+1}-1 \\right\\} \\\\\r\n& = 2 \\log 2 +0 -1 = \\underline{2 \\log 2 -1}\n\\end{align}\\]\r\n","protected":false},"excerpt":{"rendered":"\u81ea\u7136\u6570 \\(n\\) \u306b\u5bfe\u3057 \\[\\begin{align} S _ n & = \\displaystyle\\int _ 0^1 \\dfrac{1 -(-x)^n}{1+x} \\, dx \\\\ T _ n & = \\tex … \u7d9a\u304d\u3092\u8aad\u3080
\r\n\r\n\u3010 \u89e3 \u7b54 \u3011<\/h4>\r\n