1. Determin the sequence converge or diverge, and , if converges, find lim a n , where
n →∞
an =
5n 2 + 1
n 2 + 4n − 2
解答:
lim a n = lim
n →∞
n →∞
2. 若已知 a n
5+
1
n2
5n + 1
= lim
=5
n + 4 n − 2 n →∞ 1 + 4 1 − 2
n n2
2
2
) (
)(
(
= 1 − 212 1 − 312 … 1 − (n +11)2
解答:
)
n→∞
) (
)(
(
,球數列之極限 lim a n
a n = 1 − 212 1 − 312 … 1 − (n +11)2
)
(
)(
= (1 − 12 )(1 + 12 )(1 − 13 )(1 + 13 )… 1 − (n1+1) 1 + (n1+1)
利用平方差公式可得
=
n n+2 1 n+2
1 3 2 4 3
⋅ ⋅ ⋅ ⋅ …
⋅
= ⋅
2 2 3 3 4 n +1 n +1 2 n +1
1 n+2 1
lim an = lim ⋅
=
n→∞
n →∞ 2 n + 1
2
2
2
an
b
− n
n →∞ b
an
n
3. 若已知 lim a n = ∞, lim bn = ∞, lim a n − bn = k ,求 lim
n →∞
n →∞
n →∞
解答:
(
(a − bn ) an + an bb + bn
an
b
a b
− n = lim n n = lim n
n→∞ b
n →∞
a n n→∞ a n bn
a n bn
n
2
2
3
3
lim
= lim(a n − bn ) ⋅ lim
n→∞
n→∞
(a
2
n
2
+ a n bb + bn
a n bn
2
)
2
)
)
⎛a
b ⎞
= lim(an − bn ) ⋅ lim⎜⎜ n + 1 + n ⎟⎟
n →∞
n →∞ b
an ⎠
⎝ n
⎛
a
b
⎜
= k ⋅ ⎜ lim n + lim1 + lim n
n →∞ b
n →∞
an
⎜
n
n →∞
⎝
⎞
⎟
⎟ = k ⋅ (1 + 1 + 1) = 3k
⎟
⎠