Question

Conditional and absolute convergence: For each of the series that follow, determine whether the series converges absolutely, converges conditionally, or diverges. Explain the criteria you are using and why your conclusion is valid.

$\sum _{k=1}^{\infty }{(-1)}^k\frac{\sqrt{k}}{3k-1}$

Short answer

The series $\sum _{k=1}^{\infty }{(-1)}^k\frac{\sqrt{k}}{3k-1}$converges conditionally.

Step-by-step solution

Step 1. Given Information.

The series:

$\sum _{k=1}^{\infty }{(-1)}^k\frac{\sqrt{k}}{3k-1}$

Step 2. By Alternating Series Test.

According to the Alternating Series Test, the sequence $a_{k+1}<a_k$for every . Then the alternating series $a_{k+1},a_k$both converges.

Step 3. Find ak+1.

$$\begin{gathered} a_k=\frac{\sqrt{k}}{3k-1} \\ {a}_{k+1}=\frac{\sqrt{k+1}}{3(k+1)-1} \\ =\frac{\sqrt{k+1}}{3k+2} \\ {a}_{k+1}<a_k \end{gathered}$$

So the sequence is monotonic decreasing sequence.

Step 4. Find limk→∞ak.

$$\begin{gathered} \underset{k\to \infty }{lim}{a}_k=\underset{k\to \infty }{lim}\frac{\sqrt{k}}{3k-1} \\ =0 \end{gathered}$$

So the series converges.

Step 5. Use Limit Comparison test.

$$\begin{gathered} b_k=\frac{1}{\sqrt{k}} \\ \underset{k\to \infty }{lim}\frac{a_k}{b_k}=\underset{k\to \infty }{lim}\frac{{\displaystyle \frac{\sqrt{k}}{3k-1}}}{{\displaystyle \frac{1}{\sqrt{k}}}} \\ =\underset{k\to \infty }{lim}\frac{k}{3k-1} \\ =\frac{1}{3} \end{gathered}$$

So the series $b_k$diverges.

And by Limit Comparison Test, the series converges conditionally.