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.

$\sum _{k=0}^{\mathrm{\infty }} \frac{(-1{)}^{k+1}}{1+k^2}$

Short answer

The series$\sum _{k=0}^{\mathrm{\infty }} \frac{(-1{)}^{k+1}}{1+k^2}$converges

Step-by-step solution

Step 1. Given information

$\sum _{k=0}^{\mathrm{\infty }} \frac{(-1{)}^{k+1}}{1+k^2}$

Step 2. Solving the series

$\sum _{k=0}^{\mathrm{\infty }} \frac{(-1{)}^{k+1}}{1+k^2}=\sum _{k=1}^{\mathrm{\infty }} (-1{)}^{k+1}{a}_k\text{where}{a}_k=\frac{1}{1+k^2}$

Substitute $k=k+1$in$a_k=\frac{1}{1+k^2}$,

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

Step 3. Calculate the limit

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

Therefore,$\sum _{k=0}^{\mathrm{\infty }} \frac{(-1{)}^{k+1}}{1+k^2}$converges absolutely.