设正数数列{an}为一等比数列，且a2=4，a4=16．求：limn→∞

设数列{an}的公比为q，显然q≠1，a4a2＝q2＝4，由于an＞0，n∈N，∴q=2，a1＝a2q＝2，∴an=a1qn-1=2n，因此lgan+1+lgan+2+…+lga2nn=lg2n+1+lg2n+2+…+lg22nn2=[(n+1)+(n+2)+…+2n]n2lg2=3n2+n2n2•lg2，原式=limn→∞(3n2+n2n2•lg2) ＝lg2•limn→∞3n2+n2n2=32lg2．