Q. 36 - Calculus

Question

In Exercises 35–40 use the root test to analyze whether the given series converges or diverges. If the root test is inconclusive, use a different test to analyze the series.

$\sum _{k=1}^{\infty }\frac{5^k}{k^2{4}^{k+1}}$

Step-by-Step Solution

Answer

The given series is diverges.

Step 1. Given Information.

The given series is $\sum _{k=1}^{\infty }\frac{5^k}{k^2{4}^{k+1}}.$

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

By using the root test, let the general term is $a_k=\frac{5^k}{k^2{4}^{k+1}}.$

So,

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

Since $\frac{5}{4}>1,$ by using the root test the given series is diverges.