序列a(k)有界,b(k)=a(k+1)-a(k),当k趋近正无穷时,(k-2)*ln(k+1)*b(k)+a(k)趋近零.则a(k)是否趋近零?求证明.ln(k)为真数为k的自然对数.

Question

序列a(k)有界,b(k)=a(k+1)-a(k),当k趋近正无穷时,(k-2)*ln(k+1)*b(k)+a(k)趋近零.则a(k)是否趋近零?
求证明.ln(k)为真数为k的自然对数.

Answer 1

b(k)(k-2)ln(k+1)+a(k)
=[a(k+1)-a(k)](k-2)ln(k+1)+a(k)
=a(k+1)(k-2)ln(k+1)+a(k)[1-(k-2)ln(k+1)]
lim(k->+∞) a(k+1)(k-2)ln(k+1)=-lim(k->+∞) a(k)[1-(k-2)ln(k+1)]
=lim(k->+∞) a(k)[(k-2)ln(k+1)-1]
左边变为右边的分母
lim(k->+∞) a(k)[(k-2)ln(k+1)-1]/[a(k+1)(k-2)ln(k+1)]
=lim(k->+∞) [a(k)/a(k+1)]-1/[a(k+1)(k-2)ln(k+1)]
=1
因为(k-2)ln(k+1)趋向于∞
且a(k+1)有界
则lim(k->+∞) 1/[a(k+1)(k-2)ln(k+1)]=0
所以lim(k->+∞) a(k)/a(k+1)=1
说明当k->+∞时,因为a(k)有界
所以lim(k->+∞) a(k)=lim(k->+∞) a(k+1)
所以lim(k->+∞) b(k)=0
原式
lim(k->+∞) (k-2)*ln(k+1)*b(k)+a(k)=0
所以
lim(k->+∞) a(k)=0