 | 条件からa[n]は正なので
(n-1)a[n]≧0
(n-1)a[n]+1/(2^n)＞0
(2n-1)a[n]+1/(2^n)＞na[n]
{(2n-1)/(2n)}a[n]+1/{n・2^(n+1)}＞a[n]/2
左辺はa[n+1]に等しいのでa[n+1]＞a[n]/2
a[1]＞1/2なのでa[n]＞1/(2^n)
(2^n)a[n]＞1
(2^n)a[n]-1＞0
-(2^n)a[n]+1＜0
a[n+1]={(2n-1)/(2n)}a[n]+1/{n・2^(n+1)}
a[n+1]=a[n]-{1/(2n)}a[n]+1/{n・2^(n+1)}
a[n+1]-a[n]=-{1/(2n)}a[n]+1/{n・2^(n+1)}
a[n+1]-a[n]={-(2^n)a[n]+1}/{n・2^(n+1)}＜0
従って単調減少。
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