To check for absolute convergence of the series in part (a), we take the absolute value of each term, which gives Σ_n=1^(∞) (1)/(√n). This is a standard p -series of the form Σ (1)/(n^p) with p = 1/2. Because p ≤ 1, this positive-term series diverges, meaning the original series does not converge absolutely.

To check for absolute convergence of the series in part (a), we take the absolute value of each term, which gives Σ_n=1^(∞) (1)/(√n). This is a standard p -series of the form Σ (1)/(n^p) with p = 1/2. Because p ≤ 1, this positive-term series diverges, meaning the original series does not converge absolutely.

We now test the original alternating series using the Alternating Series (Leibniz) Test. The sequence of terms b_n = (1)/(√n) is strictly positive, monotonically decreasing since √(n+1) > √n, and its limit as n → ∞ is 0. Therefore, the series converges, and since its absolute series diverges, it converges conditionally.

We now test the original alternating series using the Alternating Series (Leibniz) Test. The sequence of terms b_n = (1)/(√n) is strictly positive, monotonically decreasing since √(n+1) > √n, and its limit as n → ∞ is 0. Therefore, the series converges, and since its absolute series diverges, it converges conditionally.

For part (b), taking the absolute value gives the series Σ_n=1^(∞) (n)/(n²+1). We compare this to the divergent harmonic series Σ_n=1^(∞) (1)/(n) using the Limit Comparison Test. Since the limit of their ratio as n → ∞ is 1 (a finite non-zero value), both series behave the same way, so the absolute series diverges.

For part (b), taking the absolute value gives the series Σ_n=1^(∞) (n)/(n²+1). We compare this to the divergent harmonic series Σ_n=1^(∞) (1)/(n) using the Limit Comparison Test. Since the limit of their ratio as n → ∞ is 1 (a finite non-zero value), both series behave the same way, so the absolute series diverges.

We verify the conditions of the Alternating Series Test on b_n = (n)/(n²+1). By differentiating f(x) = (x)/(x²+1), we find f'(x) = (1-x²)/((x²+1)²) 1, showing b_n is decreasing for n ≥ 1. Since lim_n→∞ b_n = 0, the series converges by Leibniz's test, which makes it conditionally convergent.

We verify the conditions of the Alternating Series Test on b_n = (n)/(n²+1). By differentiating f(x) = (x)/(x²+1), we find f'(x) = (1-x²)/((x²+1)²) 1, showing b_n is decreasing for n ≥ 1. Since lim_n→∞ b_n = 0, the series converges by Leibniz's test, which makes it conditionally convergent.

For part (c), the presence of factorials makes the Absolute Ratio Test ideal. We compute the limit of the absolute ratio | (a_n+1)/(a_n) | as n → ∞. Simplifying the factorial gives ((n+1)!)/((n+2)!) = (1)/(n+2), leading to a limit of 0. Since L = 0 < 1, the series converges absolutely.

For part (c), the presence of factorials makes the Absolute Ratio Test ideal. We compute the limit of the absolute ratio | (a_n+1)/(a_n) | as n → ∞. Simplifying the factorial gives ((n+1)!)/((n+2)!) = (1)/(n+2), leading to a limit of 0. Since L = 0 < 1, the series converges absolutely.

2.Testthefollowingseriesforconvergence. Ineachcasestatewhethertheconvergenceisabsoluteor conditional: (a) ∞ $n=1 (−1)n−1$ √n (b) ∞ $n=1 (−1)n−1n n2+1 (c)$ ∞ $n=1 (−1)n−1n3 (n+1)!$

Answer: (a) Conditionally convergent; (b) Conditionally convergent; (c) Absolutely convergent.

Step-by-step solution

Step 1: Analyze Series (a) for Absolute Convergence

To check for absolute convergence of the series in part (a), we take the absolute value of each term, which gives $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$ . This is a standard $p$ -series of the form $\sum \frac{1}{n^p}$ with $p = 1/2$ . Because $p \le 1$ , this positive-term series diverges, meaning the original series does not converge absolutely.

Step 2: Test Series (a) Using Alternating Series Test

We now test the original alternating series using the Alternating Series (Leibniz) Test. The sequence of terms $b_n = \frac{1}{\sqrt{n}}$ is strictly positive, monotonically decreasing since $\sqrt{n+1} > \sqrt{n}$ , and its limit as $n \to \infty$ is $0$ . Therefore, the series converges, and since its absolute series diverges, it converges conditionally.

Step 3: Analyze Series (b) for Absolute Convergence

For part (b), taking the absolute value gives the series $\sum_{n=1}^{\infty} \frac{n}{n^2+1}$ . We compare this to the divergent harmonic series $\sum_{n=1}^{\infty} \frac{1}{n}$ using the Limit Comparison Test. Since the limit of their ratio as $n \to \infty$ is $1$ (a finite non-zero value), both series behave the same way, so the absolute series diverges.

Step 4: Test Series (b) Using Alternating Series Test

We verify the conditions of the Alternating Series Test on $b_n = \frac{n}{n^2+1}$ . By differentiating $f(x) = \frac{x}{x^2+1}$ , we find $f'(x) = \frac{1-x^2}{(x^2+1)^2} < 0$ for all $x > 1$ , showing $b_n$ is decreasing for $n \ge 1$ . Since $\lim_{n\to\infty} b_n = 0$ , the series converges by Leibniz's test, which makes it conditionally convergent.

Step 5: Test Series (c) Using the Ratio Test

For part (c), the presence of factorials makes the Absolute Ratio Test ideal. We compute the limit of the absolute ratio $\left| \frac{a_{n+1}}{a_n} \right|$ as $n \to \infty$ . Simplifying the factorial gives $\frac{(n+1)!}{(n+2)!} = \frac{1}{n+2}$ , leading to a limit of $0$ . Since $L = 0 < 1$ , the series converges absolutely.

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