=∑(x^(2n-1))'/2^n=((1/x)∑(x²/2)^n)'=(1/x(1-x²/2))'=(1/x+x/(2-x²))'=(1/x+1/2√2(1/(√2+x)+1/(√2-x)))'=-1/x²+(1/(x-√2)²-1/(x+√2)²)/2√2
哈O(∩_∩)O哈哈~
