Question

求幂级数的和函数;并求级数的和.如图

求幂级数的和函数;并求级数的和.

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Answer 1

首先1/(1-x) = ∑{0 ≤ n} x^n, 在(-1,1)内闭一致收敛.
逐项积分得-ln(1-x) = ∑{0 ≤ n} x^(n+1)/(n+1) = ∑{1 ≤ n} x^n/n (由ln(1) = 0可确定积分常数).
于是-x·ln(1-x) = ∑{1 ≤ n} x^(n+1)/n = ∑{2 ≤ n} x^n/(n-1),
而-ln(1-x)/x = ∑{1 ≤ n} x^(n-1)/n = ∑{0 ≤ n} x^n/(n+1) = 1+x/2+∑{2 ≤ n} x^n/(n+1).
即-ln(1-x)/x-1-x/2 = ∑{2 ≤ n} x^n/(n+1).
相减得ln(1-x)/x+1+x/2-x·ln(1-x) = ∑{2 ≤ n} (1/(n-1)-1/(n+1))·x^n = 2·∑{2 ≤ n} x^n/(n²-1).
即∑{2 ≤ n} x^n/(n²-1) = ln(1-x)/(2x)+1/2+x/4-x·ln(1-x)/2, 在(-1,1)内闭一致收敛.
代入x = 1/2即得∑{2 ≤ n} 1/((n²-1)2^n) = 5/8-3ln(2)/4.