cauchy second theorem on limits proof
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Answer:
hey mate here's your ans
Step-by-step explanation:
Cauchy's second theorem on limits says that for a sequence {an} of positive values, if
limn→∞
an+1
an
=ℓ,
then
limn→∞ (an)1/n=ℓ.
You are wanting the limit limn→∞ (an)1/n with an=(
n!
nn
)n. What happens when we simplify the ratio
an+1
an
?
an+1
an
=
(
(n+1)!
(n+1)(n+1)
)n+1
(
n!
nn
)n
=
(
n!
(n+1)n
)n+1
(
n!
nn
)n
= (
n!
(n+1)n
)⋅[
(
n!
(n+1)n
)
(
n!
nn
)
]n = (
n!
(n+1)n
)⋅[(
n
n+1
)n]n.
The first factor in the product, (
n!
(n+1)n
), is bounded above by 1. The second factor, (
n
n+1
)n2, approaches 0 in the limit (hint: use L'Hôpital's rule to show n2log(n/(n+1))→−∞). Therefore,
limn→∞
an+1
an
=0.
please follow
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