设级数∑∞n=1un收敛，试判定下列级数的敛散性：
(1)∑∞n=1un+1；(2)∑∞n=1(un+10)；
(3)∑∞n=110un；(4)∑∞n=1(un/10.
(1)收敛 设∑∞n=1un的前n项和为sn，limn→∞sn=s． ∑∞n=1un+1的前n项和为6n 则 6n=u2+u3+…un+2=sn+1-u1 所以 limn→∞6n=limn→∞(sn+1-u1)=s-u1 (2)发散 设∑∞n=1un的前n项和为sn，limn→∞sn=s ∑∞n=1(un+10)的前n项和为6n 则6n=(u1+10)+(u2+10)+…+(un+10)=sn+10n 所以∑∞n=16n=limn→∞(sn+10n)=∞ (3)收敛 设∑∞n=1un的前n项和为sn-limn→∞sn=s ∑∞n=110un的前n项和为6n 则 6n=10u1+10u2+…+10un=10sn 所以 limn→∞6n=limn→∞10sn=10s (4)收敛 设∑∞n=1un的前n项和为sun，limn→∞sun=s ∑∞n=1(un/10)的前n项和为6n 则6n=u1/10+u2/10+…+un/10=sn/10 则limn→∞6n=limn→∞sn/10=s/10
