Q. 48 - 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. 

$\sum _{k=1}^{\infty }{\left(\frac{1}{k^2}\right)}^{\sqrt{k}}$

Step-by-Step Solution

Answer

The given series converges.

Step 1. Given Information.

The given series is $\sum _{k=1}^{\infty }{\left(\frac{1}{k^2}\right)}^{\sqrt{k}}.$

Step 2. Determine whether the given series converges or diverges.

We will use the root test to determine whether the given series converges or diverges, since the series has positive terms so, it meets the hypothesis of the test. 

Let the general term is $a_k={\left(\frac{1}{k^2}\right)}^{\sqrt{k}}.$

So,

$$\begin{gathered} \rho =\underset{k\to \infty }{\lim }{\left(a_k\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$k$}\right.} \\ \rho =\underset{k\to \infty }{\lim }{\left({\left(\frac{1}{k^2}\right)}^{\sqrt{k}}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$k$}\right.} \\ \rho =\underset{k\to \infty }{\lim }{\left({\left(\frac{1}{k^2}\right)}^{\frac{k}{2}}\right)}^{\frac{1}{k}} \\ \rho =\underset{k\to \infty }{\lim }\left({\left(\frac{1}{k^2}\right)}^{\frac{1}{2}}\right) \\ \rho =\underset{k\to \infty }{\lim }\left(\frac{1}{k}\right) \\ \rho =0 \end{gathered}$$

Since $0<1,$ by using the root test the given series converges.