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.

$\sum _{k=1}^{\infty }\frac{k}{3k+100}$

Short answer

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

Step-by-step solution

Step 1. Given information.

Consider the given series,

$\sum _{k=1}^{\infty }\frac{k}{3k+100}$

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$\sum _{k=1}^{\infty }\frac{k}{3k+100}$is divergent.