无穷级数∑an是发散的正项级数,Sn是前n项和,lim an/Sn=0(n趋于+∞),证明无穷级数∑an(x^n)收敛半径是1.
问题描述:
无穷级数∑an是发散的正项级数,Sn是前n项和,lim an/Sn=0(n趋于+∞),证明无穷级数∑an(x^n)收敛半径是1.
答
∵∑an发散,且Sn>an>0∴limsupan^(1/n)≥1,而liman/Sn=0 => lim(Sn-S[n-1])/Sn=0 =>limS[n-1]/Sn=1 => limsupSn^(1/n)≤limsupSn/S[n-1]=1=> limsupan^(1/n)≤limsupSn^(1/n)=1∴limsupan^(1/n)=1即级数∑anx^n的收...