Question

Use the alternating series test to determine whether the series in Exercises 24–29 converge or diverge. If a series converges, determine whether it converges absolutely or conditionally.

$\sum _{k=2}^{\mathrm{\infty }} \frac{(-1{)}^{k+1}}{(\ln k{)}^2}$

Short answer

The series$\sum _{k=2}^{\mathrm{\infty }} \frac{(-1{)}^{k+1}}{(\ln k{)}^2}$converges conditionally

Step-by-step solution

Step 1. Given information

$\sum _{k=2}^{\mathrm{\infty }} \frac{(-1{)}^{k+1}}{(\ln k{)}^2}$

Step 2. Solving the series

$\sum _{k=2}^{\mathrm{\infty }} \frac{(-1{)}^{k+1}}{(\ln k{)}^2}=\sum _{k=2}^{\mathrm{\infty }} (-1{)}^{k+1}\frac{1}{(\ln k{)}^2}$

$\sum _{k=2}^{\mathrm{\infty }} (-1{)}^{k+1}\frac{1}{(\ln k{)}^2}=\sum _{k=2}^{\mathrm{\infty }} (-1{)}^{k+1}{a}_k\text{where}{a}_k=\frac{1}{(\ln k{)}^2}$

Now, substitute $k=k+1$in $a_k=\frac{1}{(\ln k)}$,

$a_{k+1}=\frac{1}{\ln (k+1{)}^2}$

Step 3. Calculate the limit

$\begin{array}{r}\underset{k\to \mathrm{\infty }}{lim} a_k=\underset{k\to \mathrm{\infty }}{lim} \frac{1}{(\ln k{)}^2}\\ =0\end{array}$

Therefore,$\sum _{k=2}^{\mathrm{\infty }} \frac{(-1{)}^{k+1}}{(\ln k{)}^2}$converges.