Question

(1 point) Each of the following statements is an attempt to show that a given series is convergent or divergent using the Comparison Test (NOT the Limit Comparison Test). For each statement, enter Correct if the argument is valid, or enter Incorrect if any part of the argument is flawed. (Note: if the conclusion is true but the argument that led to it was wrong, you must enter Incorrect.) In(n) 1 1 In(n) Incorrect v 1. For all n > 3, , and the series diverges, so by the Comparison Test, the series E diverges. n n n n n 1 n Correct 2. For all n > 2, < 1 -, and the series converges, so by the Comparison Test, the series converges. 1-n3 1-n? TT 1 arctan(n) Correct 3. For all n > 2, arctan(n) n3 < and the series converges, so by the Comparison Test, the series Σ converges. 2n? n3 2 1 n Correct 4. For all n > 3, n < - 4 Σ Σ i Σ and the series 2Σ converges, so by the Comparison Test, the series<5, 2="" 4="" and="" the="" series="" so="" by="" comparison="" e="" converges.="" n="" n2="" -="" incorrect="" v="" 5.="" for="" all="">2, 1 n ln(n) 1 n ln(n) diverges. n 1 In(n) Correct 6. For all n > 3, In(n) n2 1 ., and the series n2 converges. n4 n2

Answer 1

(1). For all \(n \geq 3\),

\(\Rightarrow n \geq 3>e\)

\(\Rightarrow \ln _{\ln n} n>\ln _{1} e\)

\(\Rightarrow\) Since \(\sum \frac{1}{n}\) is divergent.Therefore Given Series is Divergent.

(Correct Statement.)

(2).For all \(n \geq 2\),

\(\Rightarrow 2 n^{3}>1\)

\(\Rightarrow 1-n^{3}

\(\Rightarrow \frac{1}{1-n^{3}}<\frac{1}{n^{3}} \quad \because\) LHS is -ve

\(\Rightarrow \frac{n}{1-n^{3}}<\frac{1}{n^{2}}\)

Given conditions are satisfied,(True Statement)

(3).For all \(n \geq 2\),

\(\Rightarrow \arctan (n)<\frac{\pi}{2}\)

$$ \Rightarrow \frac{\arctan (n)}{n^{3}}<\frac{\pi}{2 n^{3}} $$

Since the series,

\(\sum \frac{\pi}{2 n^{3}}\) Converges,then by Comparison test, \(\sum \frac{\arctan (n)}{n^{3}}\) also Converges.

(True Statement)

(4) For all \(n \geq 3\),

\(\Rightarrow n^{3}>8\)

\(\Rightarrow 2 n^{3}-8>n^{3}\)

\(\Rightarrow 2\left(n^{3}-4\right)>n^{3}\)

\(\Rightarrow \frac{n}{n^{3}-4}<\frac{2}{n^{2}}\)

Given conditions are satisfied,(True Statement)

(5).For all \(n \geq 2\),

\(\Rightarrow n^{2}>e\)

\(\Rightarrow 2 \ln n> 1\)

\(\Rightarrow \frac{1}{n \ln n}<\frac{2}{n}\)

Given condition is true.But given statement is wrong accoding to comparison test,(False Statement)

(6).For all \(n \geq 3\),

\(\Rightarrow n>e\)

\(\Rightarrow {\ln n} > 1\)

\(\Rightarrow \frac{\ln n}{n^{2}}>\frac{1}{n^{2}}\)

Given condition is true. But given statement is wrong accoding to comparison test,(False Statement))