Q. 23 - Calculus

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.

$\underset{k=2}{\overset{\infty }{\sum }} \frac{3k-5}{k\sqrt{k^3-4}}$.

Step-by-Step Solution

Answer

The series $\underset{k=2}{\overset{\infty }{\sum }} \frac{3k-5}{k\sqrt{k^3-4}}$ is convergent.

Step 1 . Given information

$\underset{k=2}{\overset{\infty }{\sum }} \frac{3k-5}{k\sqrt{k^3-4}}$.

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

1. If $\underset{k\to \infty }{\lim } \frac{a_k}{b_k}=L$ where $L$ is a 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 $\sum _{k=1}^{\infty } a_k$ converges.
3. If $\underset{k\to \infty }{\lim } \frac{a_k}{b_k}=\infty$, and $\sum _{k=1}^{\infty } b_k$ diverges, then $\sum _{k=1}^{\infty } a_k$ diverges.

Step 3 . The terms of the series ∑ k = 2 ∞   3 k - 5 k k 3 - 4 is positive.

Find $\sum _{k=2}^{\infty } b_k$ for the given series.

$$\begin{gathered} \sum _{k=2}^{\infty } b_k=\sum _{k=2}^{\infty } \frac{k}{k\times k^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}} \\ =\sum _{k=2}^{\infty } \frac{1}{k^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}} \end{gathered}$$

Step 4 . Next find lim k → ∞   a k b k for the given series.

$$\begin{gathered} \underset{k\to \infty }{\lim } \frac{a_k}{b_k}=\underset{k\to \infty }{\lim } \frac{{\displaystyle \frac{3k-5}{k\sqrt{k^3-4}}}}{{\displaystyle \frac{1}{k^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}}} \\ =\underset{k\to \infty }{\lim } \frac{k^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(3k-5\right)}{k\sqrt{k^3-4}} \\ =\underset{k\to \infty }{\lim } \frac{k^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times k\left(3-{\displaystyle \frac{5}{k}}\right)}{k\times k^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\sqrt{1-{\displaystyle \frac{4}{k^3}}}} \\ =\underset{k\to \infty }{\lim } \frac{\left(3-\frac{5}{k}\right)}{\sqrt{1-\frac{4}{k^3}}} \\ =3 \end{gathered}$$

Step 5 . From the obtained values,

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

The value of $\sum _{k=2}^{\infty } b_k=\sum _{k=2}^{\infty } \frac{1}{k^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$ is convergent by $p-$series test.

Therefore, $\sum _{k=2}^{\infty } a_k$ is also convergent.

Hence, the given series is convergent.