Q. 19 - Calculus

Question

Use the divergence test to analyze the given series. Each answer should either be the series diverges or the divergence test fails, along with the reason for your answer.

$\underset{k=1}{\overset{\infty }{\sum }}\frac{k}{3k+100}$

Step-by-Step Solution

Answer

The series diverges as $\underset{k\to \infty }{\lim } \frac{k}{3k+100}=\frac{1}{3}$.

Step 1. Given information.

Consider the given series,

$\underset{k=1}{\overset{\infty }{\sum }}\frac{k}{3k+100}$

Step 2. Analyze the series.

The divergence test states that if the sequence $\left\{a_k\right\}$does not converge to zero, then the series $\underset{k=1}{\overset{\infty }{\sum a_k}}$ diverges.

The value of the sequence $\left\{a_k\right\}=\left\{\frac{1}{3k+100}\right\}$ is given below,

$$\begin{gathered} \underset{k\to \infty }{\lim }{a}_k=\underset{k\to \infty }{\lim }\left(\frac{k}{3k+100}\right) \\ =\underset{k\to \infty }{\lim }\left(\frac{1}{3+{\displaystyle \frac{100}{k}}}\right) \\ =\frac{1}{3+0} \\ =\frac{1}{3} \end{gathered}$$

The series $\sum _{k=1}^{\infty } \frac{k}{3k+100}$ by divergence test is divergent as $\underset{k\to \infty }{\lim }\left(\frac{k}{3k+100}\right)=\frac{1}{3}$, which is non-zero.

Thus, the series $\underset{k=1}{\overset{\infty }{\sum }}\frac{k}{3k+100}$ is divergent.