判断无穷级数∑n=1∞1/√n+1+√n的敛散性．
∵1/√n+1+√n=(√n+1-√n)/[(√n+1+√n)(√n+1-√n)]=√n+1-√n ∴∑n=1∞1/(√n+1+√n)=∑n=1∞(√n+1-√n)=(√2-√1)+(√3-√2)+(√4-√3)+…+(√n+1-√n)+…=√n+1-1，发散．
判断无穷级数∑n=1∞1/√n+1+√n的敛散性．
∵1/√n+1+√n=(√n+1-√n)/[(√n+1+√n)(√n+1-√n)]=√n+1-√n ∴∑n=1∞1/(√n+1+√n)=∑n=1∞(√n+1-√n)=(√2-√1)+(√3-√2)+(√4-√3)+…+(√n+1-√n)+…=√n+1-1，发散．
