Q. 44 - Calculus

Question

Use any convergence test from Sections 7.4–7.6 to determine whether the series in Exercises 41–59 converge or diverge. Explain why each series that meets the hypotheses of the test you select does so.

$\sum _{k=1}^{\infty }\frac{k^3}{k^2+7}$

Step-by-Step Solution

Answer

The given series diverges.

Step 1. Given Information.

The given series is $\sum _{k=1}^{\infty }\frac{k^3}{k^2+7}.$

Step 2. Determine whether the series converges or diverges.

To determine whether the series converges or diverges we will use the divergence test.

The general term of the given series is $a_k=\frac{k^3}{k^2+7}.$

Let's calculate

$$\begin{gathered} \underset{k\to \infty }{\lim }\frac{k^3}{k^2+7} \\ =\underset{k\to \infty }{\lim }\frac{k^3}{k^2\left(1+{\displaystyle \frac{7}{k^2}}\right)} \\ =\underset{k\to \infty }{\lim }\frac{k}{\left(1+{\displaystyle \frac{7}{k^2}}\right)} \\ =\infty \end{gathered}$$

Since the series doesn't converge to zero; hence it meets the hypothesis of the Divergence test.

Therefore, according to the divergence test the given series diverges.