Question

请问这两道无穷级数怎么做？

Answer 1

1. 当n = k²时系数a[k²] = 1/2^k, 其余情况a[n] = 0.
可知limsup{n→∞} a[n]^(1/n) = lim{k→∞} a[k²]^(1/k²) = lim{k→∞} 1/2^(1/k) = 1.
根据Cauchy-Hadamard公式, 收敛半径R = 1/limsup{n→∞} a[n]^(1/n) = 1.
于是级数对x∈(-1,1)收敛.
对x = 1, 易见级数∑1/2^n收敛.
而对x = -1, 有(-1)^(n²) = (-1)^n, 因此级数∑(-1)^(n²)/2^n = ∑(-1)^n/2^n, 也易见是收敛的.
综上幂级数的收敛域为[-1,1].
2. ∑{1 ≤ n} (2n-1)·x^(2n-2)/2^n
= ∑{1 ≤ n} (x^(2n-1))'/2^n
= (∑{1 ≤ n} x^(2n-1)/2^n)'
= (∑{0 ≤ n} x^(2n+1)/2^(n+1))'
= (x/2·∑{0 ≤ n} x^(2n)/2^n)'
= (x/2·∑{0 ≤ n} (x²/2)^n)'
= ((x/2)/(1-x²/2))'
= (x/(2-x²))'
= (2+x²)/(2-x²)²