11/12/21 23:19:52.06
>>804-805
{2n/(2n+1)}e < a_n < {(2n+1)/(2n+2)}e,
{(2n+2)/(2n+1)}e < b_n < {(2n+1)/2n}e,
は同値
左側
>>811 と相加相乗平均より
{2√(n(n+1))／(2n+1)}e < e < g_n,
右側
(1 - 1/k^2)^(k+1) > 1 -(k+1)/k^2 = (k^2 -k-1)/k^2, (下に凸)
a[k]/a[k-1] = (k+1)^k･(k-1)^(k-1)／k^(2k-1) > {2k/(2k-1)}･{(2k+1)/(2k+2)},
k = n+1～∞ について掛けて
e / a[n] > (2n+2)/(2n+1),
a[n] < {(2n+1)/(2n+2)}e,