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junior college 2 | H3 Maths
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For part(b), isit divergent or convergent, and why?
Date Posted: 4 years ago
Views: 392
(3ⁿ + 5ⁿ) / 4ⁿ
Since 5ⁿ > 4ⁿ for all n ∈ Z+ ,
and 3ⁿ > 0 for all n ∈ Z+,
Then (3ⁿ + 5ⁿ) / 4ⁿ > 1 for all n ∈ Z+
Since 5ⁿ > 4ⁿ for all n ∈ Z+ ,
and 3ⁿ > 0 for all n ∈ Z+,
Then (3ⁿ + 5ⁿ) / 4ⁿ > 1 for all n ∈ Z+
So,∑(1 to ∞) (3ⁿ + 5ⁿ) / 4ⁿ > ∑(1 to ∞) 1
Since ∑(1 to ∞) 1 tends to infinity, it diverges.
Therefore ∑(1 to ∞) (3ⁿ + 5ⁿ)/4ⁿ also tends to infinity since it is bigger than ∑(1 to ∞) 1
So it is divergent.
(This is known the comparison test.)
Alternatively,
Divergence test :
If the limit of individual term ≠ 0 as n →∞,
the series diverges.
(3ⁿ + 5ⁿ)/4ⁿ = (3/4)ⁿ + (5/4)ⁿ
(5/4)ⁿ > 5/4 for all n ∈ Z+ and (5/4)ⁿ → ∞ as n → ∞
So even if (3/4)ⁿ < 1 all n ∈ Z+ and (3/4)ⁿ → 0 as n → ∞ ,
Overall,
(3/4)ⁿ + (5/4)ⁿ → ∞ as n → ∞
The limit is never 0
Since ∑(1 to ∞) 1 tends to infinity, it diverges.
Therefore ∑(1 to ∞) (3ⁿ + 5ⁿ)/4ⁿ also tends to infinity since it is bigger than ∑(1 to ∞) 1
So it is divergent.
(This is known the comparison test.)
Alternatively,
Divergence test :
If the limit of individual term ≠ 0 as n →∞,
the series diverges.
(3ⁿ + 5ⁿ)/4ⁿ = (3/4)ⁿ + (5/4)ⁿ
(5/4)ⁿ > 5/4 for all n ∈ Z+ and (5/4)ⁿ → ∞ as n → ∞
So even if (3/4)ⁿ < 1 all n ∈ Z+ and (3/4)ⁿ → 0 as n → ∞ ,
Overall,
(3/4)ⁿ + (5/4)ⁿ → ∞ as n → ∞
The limit is never 0
See 1 Answer
Divergent, since 5^n>4^n. Hence for each value of n the fraction is at least 1
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