Question

In Exercises 21–30 use one of the comparison tests to determine whether the series converges or diverges. Explain how the given series satisfies the hypotheses of the test you use.

$\sum _{k=0}^{\infty }\frac{3k^2+1}{k^3+k^2+5}$.

Short answer

The series$\sum _{k=0}^{\infty }\frac{3k^2+1}{k^3+k^2+5}$is divergent.

Step-by-step solution

Step 1. Given information

$\sum _{k=0}^{\infty }\frac{3k^2+1}{k^3+k^2+5}$.

Step 2. The comparison test states that for ∑k=1∞ak and ∑bkk=1∞ be the two series with positive terms then,

1. If $\underset{k\to \infty }{\lim }\frac{a_k}{b_k}=L$, where $L$is any positive real number, then either both converge or both diverge.
2. If $\underset{k\to \infty }{\lim }\frac{a_k}{b_k}=0$, and $\sum _{k=1}^{\infty }{b}_k$converges, then $\underset{k=1}{\overset{\infty }{\sum a_k}}$converges.
3. If $\underset{k\to \infty }{\lim }\frac{a_k}{b_k}=\infty$and $\sum _{k=1}^{\infty }{b}_k$diverges, then$\underset{k=1}{\overset{\infty }{\sum a_k}}$diverges.

Step 3. The term series of the ∑k=0∞3k2+1k3+k2+5 is positive.

Find $\underset{k=0}{\overset{\infty }{\sum b_k}}$for the given series.

$$\begin{gathered} \sum _{k=0}^{\infty }{b}_k=\sum _{k=0}^{\infty }\frac{k^2}{k^3} \\ =\sum _{k=0}^{\infty }\frac{1}{k} \end{gathered}$$

Step 4. Next find limk→∞akbkfor the given series.

$$\begin{gathered} \underset{k\to \infty }{\lim }\frac{a_k}{b_k}=\underset{k\to \infty }{\lim }\frac{{\displaystyle \frac{3k^2+1}{k^3+k^2+5}}}{{\displaystyle \frac{1}{k}}} \\ =\underset{k\to \infty }{\lim }\frac{k(3k^2+1)}{k^3+k^2+5} \\ =\underset{k\to \infty }{\lim }\frac{k^3\left(3+\frac{1}{k^2}\right)}{k^3\left(1+{\displaystyle \frac{1}{k}}+{\displaystyle \frac{1}{k^3}}\right)} \\ =\underset{k\to \infty }{\lim }\frac{\left(3+\frac{1}{k^2}\right)}{\left(1+{\displaystyle \frac{1}{k}}+{\displaystyle \frac{1}{k^3}}\right)} \\ =3 \end{gathered}$$

Step 5. From the obtained values,

The value of $\underset{k\to \infty }{\lim }\frac{a_k}{b_k}=3$which is a non zero finite number.

The value of $\sum _{k=0}^{\infty }{b}_k=\sum _{k=0}^{\infty }\frac{1}{k}$is divergent by $p-$series test.

Therefore, the series $\sum _{k=0}^{\infty }{a}_k$is also divergent.

Hence the given series is divergent.