Q. 31 - Calculus

Question

Use any convergence test from this section or the previous section to determine whether the series in Exercises $31-48$ converge or diverge. Explain how the series meets the hypotheses of the test you select.

$\sum _{k=1}^{\infty } k^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$.

Step-by-Step Solution

Answer

The series $\sum _{k=1}^{\infty } k^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ is divergent.

Step 1 . Given information

$\sum _{k=1}^{\infty } k^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$.

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

1. If $\underset{k\to \infty }{\lim } \frac{a_k}{b_k}=L$, where $L$ is any positive real number.
2. If $\underset{k\to \infty }{\lim } \frac{a_k}{b_k}=0$ and $\sum _{k=1}^{\infty } b_k$ converges, then $\sum _{k=1}^{\infty } a_k$ also converges.
3. If $\underset{k\to \infty }{\lim } \frac{a_k}{b_k}=\infty$, and $\sum _{k=1}^{\infty } b_k$ diverges, then $\sum _{k=1}^{\infty } a_k$ also diverges.

Step 3 . The terms of the series ∑ k = 1 ∞   1 k 1 2 are positive.

Find $\sum _{k=1}^{\infty } b_k$ for the given series.

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

Next find $\underset{k\to \infty }{\lim } \frac{a_k}{b_k}$ for the given series.

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

Step 4 . From the obtained values,

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

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

Therefore, the series $\sum _{k=1}^{\infty } a_k$ is also divergent.

Hence, the given series is divergent.