Question

Use any convergence tests from Sections 7.4–7.7 to determine whether the series in Exercises 36–51 converge absolutely, converge conditionally, or diverge. Explain why the series meets the hypotheses of the test you select.
$\sum _{k=1}^{\infty }\frac{{(-1)}^{k}k}{k+1}$

Short answer

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

Step-by-step solution

Step 1. Given Information.

The series:

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

Step 2. Find limk→∞ak.

$$\begin{gathered} \underset{k\to \infty }{lim}\left|a_k\right|=\underset{k\to \infty }{lim}\left|\frac{{(-1)}^{k}k}{k+1}\right| \\ =\underset{k\to \infty }{lim}\left|\frac{k}{k+1}\right| \\ =\underset{k\to \infty }{lim}\left|\frac{k}{k(1+{\displaystyle \frac{1}{k}})}\right| \\ =1 \end{gathered}$$

Step 3. By Divergence test.

By Divergence test, the series $\underset{k\to \infty }{lim}\left|a_k\right|\ne 0$.

So the series is divergent.