首先看∑1/ln(1+n)因为lim(n→∞)1/ln(1+n)/(1/n)=lim(n→∞)n/ln(1+n)=lim(n→∞)1/(1/(n+1))=lim(n→∞)n+1=∞而∑1/n发散,所以∑1/ln(1+n)发散所以不是绝对收敛然后对于交错级数∑(-1)^n-1/ln(1+n)收敛性,由莱布里茨判别法:lim(n→∞)1/ln(1+n)=0且1/ln(1+n)>1/ln(n+2)所以交错级数∑(-1)^n-1/ln(1+n)收敛,且和S