幂级数求收敛域内的和函数 下面该怎么做啊
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幂级数求收敛域内的和函数 下面该怎么做啊
![幂级数求收敛域内的和函数 下面该怎么做啊](/uploads/image/z/16055257-49-7.jpg?t=%E5%B9%82%E7%BA%A7%E6%95%B0%E6%B1%82%E6%94%B6%E6%95%9B%E5%9F%9F%E5%86%85%E7%9A%84%E5%92%8C%E5%87%BD%E6%95%B0+%E4%B8%8B%E9%9D%A2%E8%AF%A5%E6%80%8E%E4%B9%88%E5%81%9A%E5%95%8A)
∑{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/(1-x²/2))'
= (x/(2-x²))'
= (2+x²)/(2-x²)².
代入x = 1得∑{1 ≤ n} (2n-1)/2^n = 3.
= ∑{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/(1-x²/2))'
= (x/(2-x²))'
= (2+x²)/(2-x²)².
代入x = 1得∑{1 ≤ n} (2n-1)/2^n = 3.