Q. 39 - 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.

$\sum _{k=1}^{\infty }{\left(\frac{1+4k}{5k-1}\right)}^k$

Step-by-Step Solution

Answer

The given series converges.

Step 1. Given Information.

The given series is $\sum _{k=1}^{\infty }{\left(\frac{1+4k}{5k-1}\right)}^k.$

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

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

So,

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

Since $\frac{4}{5}<1,$ by using the root test the given series converge.