设S(x)=∑(n=1→∞)x^(2n-1)/(2n-1)
则S'(x)=∑(n=0→∞)x^(2n)=∑(n=0→∞)(x²)^n=1/(1-x²),x∈(-1,1)
S(x)=∫dx/(1-x²)=1/2*ln|(x+1)/(x-1)|+C
而S(0)=0,代入上式得C=0,∴S(x)=1/2*ln|(x+1)/(x-1)|
而1/2^n=(1/√2)^(2n)=1/√2*(1/√2)^(2n-1)
∴原式=1/√2*1/2*ln|(1/√2+1)/(1/√2-1)|=√2/4*ln(3+2√2)
追问答案是√2/2*ln(1+√2)诶
答案是√2/2*ln(1+√2)诶
追答你先看看和函数对了没
追问嗯嗯对的,看错了谢谢,还提问了两道题麻烦看一下