Question

Use any convergence tests to determine whether the series converge absolutely, converge conditionally, or diverge. Explain why the series meets the hypotheses of the test you select.

$\sum _{k=1}^{\infty }{\left(-1\right)}^k\frac{\sqrt[3]{k^2+3k+1}}{k+1}$

Short answer

The series converges conditionally.

Step-by-step solution

Step 1. Given information.

Consider the given question,

$\sum _{k=1}^{\infty }{\left(-1\right)}^k\frac{\sqrt[3]{k^2+3k+1}}{k+1}$

Step 2. Use the alternating series test.

Using the alternating series test, assume $\left\{a_k\right\}$be the sequence of positive numbers.

If $a_{k+1}<a_k$for every $k\ge 1$and $\underset{k\to \infty }{\lim }{a}_k=0$

Then the alternating series $\sum _{k=1}^{\infty }{\left(-1\right)}^{k+1}{a}_k$and $\sum _{k=1}^{\infty }{\left(-1\right)}^k{a}_k$both converge.

Consider the series $\sum _{k=1}^{\infty }{\left(-1\right)}^k{a}_k=\sum _{k=1}^{\infty }\frac{{\left(-1\right)}^k\sqrt[3]{k^2+3k+1}}{k+1}$.

The sequence $\left\{a_k\right\}=\left\{\frac{\sqrt[3]{k^2+3k+1}}{k+1}\right\}$is of positive terms.

Here, clearly the series is decreasing as$a_{k+1}<a_k$.

Step 3. Find the value of limit.

The value of $\underset{k\to \infty }{\lim }{a}_k$is given below,

$$\begin{gathered} \underset{k\to \infty }{\lim }{a}_k=\underset{k\to \infty }{\lim }\left(\frac{\sqrt[3]{k^2+3k+1}}{k+1}\right) \\ =0 \end{gathered}$$

Since all the conditions satisfied. Then, by alternating series test, the series is convergent.

As,

$$\begin{gathered} \sum _{k=1}^{\infty }\left|\frac{{\left(-1\right)}^k\sqrt[3]{k^2+3k+1}}{k+1}\right|=\sum _{k=1}^{\infty }\left|\frac{\sqrt[3]{k^2+3k+1}}{k+1}\right| \\ =\sum _{k=1}^{\infty }\frac{\sqrt[3]{k^2+3k+1}}{k+1} \end{gathered}$$

Step 4. Consider the dominant term of numerator and denominator.

The terms of the series $\sum _{k=1}^{\infty }\frac{\sqrt[3]{k^2+3k+1}}{k+1}$ are positive.

The series $\sum _{k=1}^{\infty }{b}_k$for the series $\sum _{k=1}^{\infty }\frac{\sqrt[3]{k^2+3k+1}}{k+1}$is given by,

$$\begin{gathered} \sum _{k=1}^{\infty }{b}_k=\sum _{k=1}^{\infty }\frac{k^{{\displaystyle \frac{2}{3}}}}{k} \\ =\sum _{k=1}^{\infty }\frac{1}{k^{{\displaystyle \frac{1}{3}}}} \end{gathered}$$

Step 5. Find the ratio of the limit.

The ratio $\underset{k\to \infty }{\lim }\frac{a_k}{b_k}$is given by,

$$\begin{gathered} \underset{k\to \infty }{\lim }\frac{a_k}{b_k}=\underset{k\to \infty }{\lim }\frac{{\displaystyle \frac{\sqrt[3]{k^2+3k+1}}{k+1}}}{{\displaystyle \frac{1}{k^{{\displaystyle \frac{1}{3}}}}}} \\ =\underset{k\to \infty }{\lim }\frac{k^{{\displaystyle \frac{1}{3}}}\times \sqrt[3]{k^2+3k+1}}{k+1} \\ =\underset{k\to \infty }{\lim }\frac{k^{\frac{1}{3}}\times k^{{\displaystyle \frac{2}{3}}}\times \sqrt[3]{1+{\displaystyle \frac{3}{k}}+{\displaystyle \frac{1}{k^2}}}}{k+1} \\ =\underset{k\to \infty }{\lim }\frac{\sqrt[3]{1+\frac{3}{k}+\frac{1}{k^2}}}{1+{\displaystyle \frac{1}{k}}} \\ =1 \end{gathered}$$

Step 6. Find the value of the limit which is non-zero finite number.

The value of $\underset{k\to \infty }{\lim }\frac{a_k}{b_k}=1$; which is non-zero finite number.

The series $\sum _{k=1}^{\infty }{b}_k=\sum _{k=1}^{\infty }\frac{1}{k^{{\displaystyle \frac{1}{3}}}}$ is divergent by p-series test.

Thus, the series $\sum _{k=1}^{\infty }{a}_k$is also divergent.

Hence, the series $\sum _{k=1}^{\infty }\frac{\sqrt[3]{k^2+3k+1}}{k+1}$is divergent.

Therefore, the given series is convergent conditionally.