设极限limn→∞∣an+1/an∣=2,则幂级数∑n=0∞anx2n的收敛区间为____.
【正确答案】:(-1/√2,1/√2)。分析:任取x0(x0≠0)点,考虑级数∑n=1∞anx02n由于limn→∞∣an+1x02n+2∣limn→∞∣an+1/an∣•x02=2x02所以2x02﹤1时收敛,2x02﹥1时发散,因此收敛区间为(-(1/√2),1/√2)
设极限limn→∞∣an+1/an∣=2,则幂级数∑n=0∞anx2n的收敛区间为____.
- 2024-08-03 20:15:07
- 高等数学(工本)(00023)
- 1