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.

$\sum _{k=1}^{\infty }\frac{k}{3k^2+100}$

Short answer

The divergence test failed as $\underset{k\to \infty }{\lim }\left(\frac{k}{3k^2+100}\right)=0$.

Step-by-step solution

Step 1. Given information.

Consider the given series,

$\sum _{k=1}^{\infty }\frac{k}{3k^2+100}$

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 $\left\{a_k\right\}=\left\{\frac{1}{\mathit{3}{k}^2\mathit{+}\mathit{100}}\right\}$is given below,

localid="1649078920875" $$\begin{gathered} \underset{k\to \infty }{\lim }{a}_k=\underset{k\to \infty }{\lim }\left(\frac{k}{3k^2+100}\right) \\ =\underset{k\to \infty }{\lim }\left(\frac{{\displaystyle \frac{1}{k}}}{3+{\displaystyle \frac{100}{k^2}}}\right) \\ =\frac{0}{3+0} \\ =0 \end{gathered}$$

Thus, the divergence test failed as$\underset{k\to \infty }{\lim }\left(\frac{k}{3k^2+100}\right)=0$.