則
lim |
n→∞ |
an |
bn |
lim |
n→∞ |
a1 +(n?1)d1 |
b1 +(n?1)d 2 |
lim |
n→∞ |
| ||||
|
d1 |
d2 |
∴
lim |
n→∞ |
a1+a2+…+an |
nb2n |
lim |
n→∞ |
na1 +
| ||
n([b1 +(2n?1)d2] |
| ||
2d2 |
3 |
4 |
故答案為
3 |
4 |
lim |
n→∞ |
an |
bn |
lim |
n→∞ |
a1+a2+…+an |
nb2n |
lim |
n→∞ |
an |
bn |
lim |
n→∞ |
a1 +(n?1)d1 |
b1 +(n?1)d 2 |
lim |
n→∞ |
| ||||
|
d1 |
d2 |
lim |
n→∞ |
a1+a2+…+an |
nb2n |
lim |
n→∞ |
na1 +
| ||
n([b1 +(2n?1)d2] |
| ||
2d2 |
3 |
4 |
3 |
4 |