证明∑1to无穷{n*ln[(2n+1)/(2n-1)]-1}收敛

Question

证明∑【1 to 无穷】{n*ln[(2n+1)/(2n-1)]-1}收敛

Answer 1

ln((2n+1)/(2n-1))=ln(1+2/(2n-1))=2/(2n-1)-2/(2n-1)^2+O(1/n^3)当n充分大时
n*ln(2n+1/2n-1)-1=2n/(2n-1) -2n/(2n-1)^2 -1 +O(1/n^2)
=1/(2n-1)-2n/(2n-1)^2+O(1/n^2)=-1/(2n-1)^2+O(1/n^2)
(利用 ln(1+x)=x-x^2/2+O(x^3) x趋于0)
又因 级数1/n^2求和收敛
所以原级数收敛