Question

Convergence or divergence of a series: For each of the series that follow, determine whether the series converges or diverges. Explain the criteria you are using and why your conclusion is valid.

$\sum _{k=1}^{\infty }\ln \left(\frac{1}{k^2}\right)$

Short answer

By divergence test, the given sequence diverges.

Step-by-step solution

Step 1. Given Information

The given sequence is$\sum _{k=1}^{\infty }\ln \left(\frac{1}{k^2}\right)$.

Step 2. Apply Divergent test

• The divergence test states that if the sequence $a_k$does not converge to 0, the series $\sum _{k=1}^{\infty }{a}_k$diverges.
• In the given sequence, $a_k=\ln \left(\frac{1}{k^2}\right)$.
• Find the value of $\underset{k\to \infty }{\lim }{a}_k$.

$\begin{array}{rcl}\underset{k\to \infty }{\lim }{a}_k& =& \underset{k\to \infty }{\lim }\ln \left(\frac{1}{k^2}\right)\\ & =& \ln \left(\frac{1}{{\infty }^2}\right)\\ & =& \ln \left(0\right)\\ & =& -\infty \end{array}$

• Thus, by divergence test, the given sequence diverges.