Description
Attached you will find 10 Questions. Please, solve them with steps using calculus 2 methods.
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Explanation & Answer
Well, please read the solutions and tell me if something is unclear.
Estimate the general term:
|
5
The series ∑ 7𝑛
1 + 4 sin(3𝑛)
|≤
1+4
5
=
.
7𝑛
7𝑛
√𝑛 + 7𝑛
is (absolutely) convergent, so we can apply the direct comparison test
and state that the initial series converges absolutely.
Again, estimate the general term:
5𝑘 2 + 1
5𝑘 2 + 𝑘 2
6𝑘 2
6
|
|≤
= 4 = 2.
𝑘
𝑘
√3𝑘 8 − 2𝑘
√3𝑘 8 − 2𝑘 8
6
The series ∑ 𝑘 2 is (absolutely) convergent, so we can apply the direct comparison test
and state that the initial series converges absolutely.
Apply the root test:
(3𝑛 − 1)2
9𝑛2 − 6𝑛 + 1 9 3
√|𝑎𝑛 | =
=
→ = > 1.
2(3𝑛2 + 1)
6𝑛2 + 2
6 2
𝑛
Because the limit exists and is greater than 1, so the limit superior. Therefore, the series is
divergent.
It is an alternating series with the absolute value of the general term monotone tending to
zero:
1
𝑎𝑛 = (−1)𝑛+1 𝑏𝑛 , 𝑏𝑛 =
→ 0 and 𝑏𝑛+1 < 𝑏𝑛 .
5𝑛 − 2
Thus, the series converges by the alternating series test.
1
It does not converge absolutely because |𝑎𝑛 | = 𝑏𝑛 ≥ 5𝑛, which is divergent (apply the
direct comparison test here).
The answer is: the series is conditionally convergent.
Estimate
|
3
−
2
𝑛−1
√3𝑛5 + 1
|≤
𝑛
√3𝑛5
=
1
√3
3
𝑛 −2 .
The series ∑ 𝑛 is convergent, therefore the given series is also (absolutely) convergent
by the direct comparison test.
5
The limit of the general term is obviously ln (3), which is nonzero. Therefore...