{"id":1164,"date":"2025-07-13T14:29:29","date_gmt":"2025-07-13T05:29:29","guid":{"rendered":"https:\/\/math-friend.com\/?p=1164"},"modified":"2025-08-01T09:46:55","modified_gmt":"2025-08-01T00:46:55","slug":"%e3%80%90%e7%a5%9e%e6%88%b8%e5%a4%a7%e5%ad%a6%e5%85%a5%e8%a9%a6%e3%80%91%e3%81%82%e3%82%8b%e6%95%b0%e5%88%97%e3%81%ae%e5%b0%8f%e6%95%b0%e9%83%a8%e5%88%86%e3%81%8c%e4%b8%80%e8%87%b4%e3%81%97%e3%81%aa","status":"publish","type":"post","link":"https:\/\/math-friend.com\/?p=1164","title":{"rendered":"\u3010\u795e\u6238\u5927\u5b66\u5165\u8a66\u3011\u3042\u308b\u6570\u5217\u306e\u5c0f\u6570\u90e8\u5206\u304c\u4e00\u81f4\u3057\u306a\u3044\u3053\u3068\u306e\u8a3c\u660e(2025)"},"content":{"rendered":"\n

\u4eca\u56de\u306f\u3053\u3061\u3089\u306e\u554f\u984c\u3092\u89e3\u3044\u3066\u3044\u304d\u307e\u3059.<\/p>\n\n\n\n

\u5b9f\u6570\\(a\\)\u306b\u5bfe\u3057\u3066, \\(a\\)\u3092\u8d85\u3048\u306a\u3044\u6700\u5927\u306e\u6574\u6570\u3092\\(m\\)\u3068\u3059\u308b\u3068\u304d, \\(a-m\\)\u3092\\(a\\)\u306e\u5c0f\u6570\u90e8\u5206\u3068\u3044\u3046. \u81ea\u7136\u6570\\(n\\)\u306b\u5bfe\u3057\u3066\\(a_n=\\sqrt{n^2+1}\\)\u3068\u3057, \\(a_n\\)\u306e\u5c0f\u6570\u90e8\u5206\u3092\\(b_n\\)\u3068\u3059\u308b. \u4ee5\u4e0b\u306e\u554f\u3044\u306b\u7b54\u3048\u3088.

(1) \\(a_n<n+1\\)\u3067\u3042\u308b\u3053\u3068\u3092\u793a\u305b.
(2) \\(b_n\\)\u3092\\(n\\)\u306e\u5f0f\u3067\u8868\u305b.
(3) \u81ea\u7136\u6570\\(n\\), \\(m\\)\u306b\u5bfe\u3057\u3066, \\(n\\neq m\\)\u306a\u3089\u3070, \\(b_n\\neq b_m\\)\u3067\u3042\u308b\u3053\u3068\u3092\u793a\u305b.
(2025 \u795e\u6238\u5927\u5b66\u6587\u7cfb[2])<\/span><\/p>\n\n\n\n

\u3053\u3061\u3089(1), (2)\u306f\u7c21\u5358\u306b\u7d42\u308f\u308b\u306e\u3067\u3059\u304c, (3)\u306f\u306a\u304b\u306a\u304b\u96e3\u3057\u3044\u3067\u3059. \\(a_n=\\sqrt{n^2+1}\\)\u306f\\(n\\)\u3092\u5927\u304d\u304f\u3059\u308b\u3068, \\(n\\)\u306b\u8fd1\u3065\u304d\u305d\u3046\u3067\u3042\u308b\u3068\u3044\u3046\u76f4\u611f\u304b\u3089, \\(b_n\\)\u304c\u5358\u8abf\u6e1b\u5c11\u6570\u5217\u3067\u3042\u308d\u3046\u3053\u3068\u304c\u4e88\u60f3\u3067\u304d\u307e\u3059. \u3088\u3063\u3066\u3053\u3053\u3067\u306f, \\(n<m\\)\u306e\u3068\u304d, \\(b_n>b_m\\)\u3067\u3042\u308b\u3053\u3068\u3092\u793a\u3059\u3053\u3068\u304c\u30df\u30bd\u306b\u306a\u308a\u307e\u3059.

\u4eca\u56de\u306f\u6587\u7cfb\u306e\u554f\u984c\u3068\u3044\u3046\u3053\u3068\u3067, \u89e3\u7b54\u306b\u306f\u6570\u2162\u3067\u5b66\u3076\u30eb\u30fc\u30c8\u306e\u5165\u3063\u305f\u95a2\u6570\u306e\u5fae\u5206\u3092\u7528\u3044\u307e\u305b\u3093\u3067\u3057\u305f. \u3053\u308c\u304c\u4f7f\u3048\u308c\u3070\u4f7f\u3048\u308c\u3070(3)\u3082\u7c21\u5358\u306b\u89e3\u304f\u3053\u3068\u304c\u3067\u304d\u308b\u305f\u3081\u5225\u89e3\u3067\u305d\u306e\u89e3\u6cd5\u3092\u7d39\u4ecb\u3092\u3057\u307e\u3059.

\u305d\u308c\u3067\u306f\u89e3\u3044\u3066\u3044\u304d\u307e\u3057\u3087\u3046.<\/p>\n\n\n\n

\n

(1) 0<\\(a_n\\)\u306a\u306e\u3067,
$$
a_n<n+1 \\iff {a_n}^2<(n+1)^2
$$\u3067\u3042\u308b.
$$
(n+1)^2-{a_n}^2=n^2+2n+1-(n^2+1)=2n>0
$$\u3088\u308a,
$$
{a_n}^2<(n+1)^2
$$\u304c\u793a\u3055\u308c, \u5148\u306e\u540c\u5024\u95a2\u4fc2\u3088\u308a,
$$
a_n<n+1
$$\u3067\u3042\u308b.<\/p>\n\n\n\n

(2)
$$
n=\\sqrt{n^2}<\\sqrt{n^2+1}=a_n
$$\u3088\u308a, \\(a_n>n\\)\u3067\u3042\u308b. \u3053\u308c\u3068(1)\u3088\u308a\\(a_n\\)\u306f
$$
n<a_n<n+1
$$\u3092\u6e80\u305f\u3057\u3066\u3044\u308b. \\(n\\), \\(n+1\\)\u306f\u5171\u306b\u81ea\u7136\u6570\u3067\u3042\u308a, \u3053\u306e\u4e0d\u7b49\u5f0f\u304b\u3089\\(n\\)\u306f\\(a_n\\)\u3092\u8d85\u3048\u3066\u304a\u3089\u305a, \u307e\u305f, \\(n+1\\)\u306f\\(a_n\\)\u3092\u8d85\u3048\u3066\u3044\u308b\u3053\u3068\u304b\u3089, \\(a_n\\)\u3092\u8d85\u3048\u306a\u3044\u6700\u5927\u306e\u81ea\u7136\u6570\u306f\\(n\\)\u3067\u3042\u308b. \u3088\u3063\u3066,
$$
b_n=a_n-n=\\sqrt{n^2+1}-n
$$\u3067\u3042\u308b.

(3) \\(b_n\\)\u3092\u4e0d\u7b49\u5f0f\u8a55\u4fa1\u3059\u308b.
$$
\\begin{align}
b_n&=\\sqrt{n^2+1}-n=\\frac{\\left(\\sqrt{n^2+1}-n\\right)\\left(\\sqrt{n^2+1}+n\\right)}{\\sqrt{n^2+1}+n}\\\\[1.5ex]
&=\\frac{(n^2+1)-n^2}{\\left(\\sqrt{n^2+1}+n\\right)}=\\frac{1}{\\left(\\sqrt{n^2+1}+n\\right)}\\\\[1.5ex]
&=\\frac{1}{a_n+n}
\\end{align}
$$\u3067\u3042\u308a, \\(n<a_n<n+1\\)\u3067\u3042\u308b\u304b\u3089,
$$
\\begin{align}
n<a_n<n+1 &\\iff n+n<a_n+n<(n+1)+n \\\\[1.5ex]
&\\iff 2n<a_n+n<2n+1\\\\[1.5ex]
& \\iff \\frac{1}{2n+1}<\\frac{1}{a_n+n}<\\frac{1}{2n} \\\\[1.5ex]
&\\iff \\frac{1}{2n+1}<b_n<\\frac{1}{2n}
\\end{align}
$$\u304c\u308f\u304b\u308b.

\u3053\u306e\u4e0d\u7b49\u5f0f\u3092\u7528\u3044\u3066, \\(n<m\\)\u306e\u3068\u304d, \\(b_n>b_m\\)\u3067\u3042\u308b\u3053\u3068\u3092\u793a\u3059. \u5148\u306e\u4e0d\u7b49\u5f0f\u306f\\(m\\)\u3067\u3082\u6210\u308a\u7acb\u3064\u304b\u3089,
\\begin{align}
\\frac{1}{2m+1}<b_m<\\frac{1}{2m}
\\end{align}
\u3067\u3042\u308b. \u3088\u3063\u3066, \\(n<m\\)\u306e\u3068\u304d,
$$
\\frac{1}{2m}<\\frac{1}{2n+1}\\,\\,\u30fb\u30fb\u30fb\u2460
$$\u3092\u793a\u305b\u3070,
$$
b_m<\\frac{1}{2m}<\\frac{1}{2n+1}<b_n
$$\u3068\u306a\u308a, \\(b_n>b_m\\)\u3067\u3042\u308b\u3053\u3068\u304c\u3044\u3048\u308b. \u305d\u3057\u3066\u5b9f\u969b, \u4e0d\u7b49\u5f0f\u2460\u306f\u4ee5\u4e0b\u306e\u3088\u3046\u306b\u8a3c\u660e\u3067\u304d\u308b. \u305f\u3060\u3057, \u4ee5\u4e0b\u306e\u8a3c\u660e\u306e\u4e2d\u3067\\(n<m\\)\u3067\\(n\\), \\(m\\)\u306f\u5171\u306b\u81ea\u7136\u6570\u3067\u3042\u308b\u3053\u3068\u304b\u3089, \\(m-n\\geq 1\\)\u3092\u7528\u3044\u3066\u3044\u308b.
$$
\\begin{align}
\\frac{1}{2n+1}-\\frac{1}{2m}&=\\frac{2m-(2n+1)}{2m(2n+1)}\\\\[1.5ex]
&=\\frac{2(m-n)-1}{2m(2n+1)}\\geq \\frac{1}{2m(2n+1)}>0
\\end{align}
$$ \u3088\u3063\u3066, \u793a\u3059\u3079\u304d\u4e0d\u7b49\u5f0f\u304c\u793a\u3055\u308c\u305f.<\/p>\n<\/div><\/div>\n\n\n\n

\u7b54\u3048\u304c\u308f\u304b\u308c\u3070\u7c21\u5358\u306b\u898b\u3048\u307e\u3059\u304c, \\(b_n\\)\u306e\u4e0d\u7b49\u5f0f\u8a55\u4fa1\u306b\u6c17\u3065\u304f\u306e\u306f\u306a\u304b\u306a\u304b\u96e3\u3057\u3044\u306e\u3067\u306f\u306a\u3044\u304b\u3068\u601d\u3044\u307e\u3059. \\(n\\)\u304c\u5927\u304d\u304f\u306a\u308b\u3068, \\(\\sqrt{n^2+1}\\)\u306f\\(n\\)\u306b\u8fd1\u3065\u304d\u305d\u3046\u3067\u3042\u308b\u3068\u3044\u3046\u76f4\u611f\u3084, \\(a_1=\\sqrt{2}-1= 0.414\\cdots\\), \\(a_2=\\sqrt{5}-2= 0.236\\cdots\\), \\(a_3=\\sqrt{10}-3=0.14\\cdots\\), \u306a\u3069\u306e\u5177\u4f53\u7684\u306a\u8a08\u7b97\u7d50\u679c\u304b\u3089, \\(b_n\\)\u304c\u5358\u8abf\u6e1b\u5c11\u3057\u305d\u3046\u3067\u3042\u308b\u3068\u3044\u3046\u3053\u3068\u306f\u3059\u3050\u306b\u6c17\u3065\u3044\u3066\u307b\u3057\u3044\u3067\u3059. \u305d\u3053\u306b\u6c17\u3065\u3051\u3070, \u306a\u3093\u3068\u304b\u5f0f\u5909\u5f62\u3057\u3066\u5148\u306e\u56de\u7b54\u306b\u8fbf\u308a\u7740\u3051\u308b\u304b\u3082\u3057\u308c\u307e\u305b\u3093.

\u306a\u304a, \u3053\u306e\u6570\u5217\\(\\{a_n\\}\\)\u304c\u5358\u8abf\u6e1b\u5c11\u3057\u305d\u3046\u3067\u3042\u308b\u3053\u3068\u306b\u6c17\u3065\u3051\u3070, \u6570\u2162\u3092\u5b66\u3093\u3060\u7406\u7cfb\u306f\u4ee5\u4e0b\u306e\u3088\u3046\u306a\u5225\u89e3\u3092\u601d\u3044\u3064\u304f\u3067\u3057\u3087\u3046.<\/p>\n\n\n\n

\\(x>0\\)\u3068\u3057\u3066, \\(f(x)=\\sqrt{x^2+1}-x\\)\u3068\u304a\u304f. \\(x=\\sqrt{x^2}<\\sqrt{x^2+1}\\)\u3067\u3042\u308b\u304b\u3089,
$$
f^\\prime (x)=\\frac{x}{\\sqrt{x^2+1}}-1=\\frac{x-\\sqrt{x^2+1}}{\\sqrt{x^2+1}}<0
$$\u3068\u306a\u308a, \\(f(x)\\)\u306f\\(x>0\\)\u3067\u5358\u8abf\u6e1b\u5c11\u3067\u3042\u308b.

\u3053\u3053\u3067, \\(b_n=f(n)\\)\u3067\u3042\u308b\u304b\u3089, \\(n<m\\)\u306e\u3068\u304d, \\(b_n=f(n)>f(m)=b_m\\)\u3067\u3042\u308b. \u3088\u3063\u3066, \\(n\\neq m\\)\u306e\u3068\u304d, \\(b_n\\neq b_m\\)\u3067\u3042\u308b.<\/p>\n\n\n\n

youtube\u3067\u3082\u89e3\u8aac\u3057\u3066\u3044\u307e\u3059.<\/p>\n\n\n\n