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.

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

Short answer

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

Step-by-step solution

Step 1. Given Information.

The series:

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

Step 2. By Alternating Series Test.

According to the Alternating Series Test, the sequence $a_{k+1}<a_k$for every $k\ge 1,\underset{k\to \infty }{lim}{a}_k=0$. Then the alternating series $a_k,a_{k+1}$both converges.

Step 3. Find ak+1.

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

So the sequence is monotonic decreasing sequence.

Step 4. Find limk→∞ak.

$$\begin{gathered} \underset{k\to \infty }{lim}{a}_k=\underset{k\to \infty }{lim}\frac{1}{\sqrt{k(k+1)}} \\ =0 \end{gathered}$$

So the series converges.

Step 5. Find bk.

To find if the series converges absolutely or conditionally.

$b_k=\frac{1}{k}$which is a dominant term.

$$\begin{gathered} \underset{k\to \infty }{lim}\frac{a_k}{b_k}=\underset{k\to \infty }{lim}\frac{{\displaystyle \frac{1}{\sqrt{k(k+1)}}}}{{\displaystyle \frac{1}{k}}} \\ =\underset{k\to \infty }{lim}\frac{k}{\sqrt{k^2(1+{\displaystyle \frac{1}{k}})}} \\ =\underset{k\to \infty }{lim}\frac{1}{\sqrt{1+{\displaystyle \frac{1}{k}}}} \\ =0 \end{gathered}$$

So the series $b_k$diverges.

And by Limit Comparison Test, the series converges conditionally.