第1个回答 2011-01-16
解:为了就算方便,令A=(1/n)[(n+1)(n+2)(n+3).......(n+n)]^(1/n)
则 A=[(n+1)(n+2)(n+3).......(n+n)/n^n]^(1/n)
={[(n+1)/n][(n+2)/n][(n+3)/n]........[(n+n)/n]}^(1/n)
=[(1+1/n)(1+2/n)(1+3/n)........(1+n/n)]^(1/n)
∴lnA=(1/n)[ln(1+1/n)+ln(1+2/n)+ln(1+3/n)+........+ln(1+n/n)] (两边取自然对数 )
==>ln[lim(n->∞)A]=lim(n->∞)(lnA) (应用对数函数的连续性)
=lim(n->∞){(1/n)[ln(1+1/n)+ln(1+2/n)+ln(1+3/n)+........+ln(1+n/n)]}
=∫(0,1)ln(1+x)dx (根据定积分定义得,符号∫(0,1)表示从0到1积分)
=[xln(1+x)]│(0,1)-∫(0,1)xdx/(1+x) (应用分部积分法)
=ln2-∫(0,1)[1-1/(1+x)]dx
=ln2-[x-ln(1+x)]│(0,1)
=ln2-(1-ln2)
=2ln2-1
=ln4-lne
=ln(4/e)
==>lim(n->∞)A=4/e (两边取反自然对数)
故 lim(n->∞){(1/n)[(n+1)(n+2)(n+3).......(n+n)]^(1/n)}=4/e。本回答被提问者采纳