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.

$\sum _{k=0}^{\mathrm{\infty }} \frac{(-2{)}^{k+1}}{1+3^k}$

Short answer

The series$\sum _{k=0}^{\mathrm{\infty }} \frac{(-2{)}^{k+1}}{1+3^k}$is converging.

Step-by-step solution

Step 1. Given information

$\sum _{k=0}^{\mathrm{\infty }} \frac{(-2{)}^{k+1}}{1+3^k}$

Step 2. Solving the series

$\sum _{k=0}^{\mathrm{\infty }} \frac{(-2{)}^{k+1}}{1+3^k}=\sum _{k=0}^{\mathrm{\infty }} (-1{)}^{k+1}\frac{(2{)}^{k+1}}{1+3^k}$

$\sum _{k=0}^{\mathrm{\infty }} (-1{)}^{k+1}\frac{(2{)}^{k+1}}{1+3^k}=\sum _{k=1}^{\mathrm{\infty }} (-1{)}^{k+1}{a}_k\text{where}{a}_k=\frac{(2{)}^{k+1}}{1+3^k}$

Now, substitute $k=k+1$in $a_k=\frac{(2{)}^{k+1}}{1+3^k}$

$a_{k+1}=\frac{(2{)}^{k+1+2}}{1+3^{k+1}}$

Step 3. Calculate the limit

$\begin{array}{r}\underset{k\to \mathrm{\infty }}{lim} a_k=\underset{k\to \mathrm{\infty }}{lim} \frac{(2{)}^k}{1+3^k}\\ =\underset{k\to \mathrm{\infty }}{lim} \frac{\frac{(2{)}^k}{(3{)}^k}}{(3{)}^k+1}\\ =\underset{k\to \mathrm{\infty }}{lim} \frac{{\left(\frac{1}{3}\right)}^k}{(3{)}^k+1}\\ \left.=\frac{\underset{k\to \mathrm{\infty }}{lim} {\left(\frac{2}{3}\right)}^k}{\underset{k\to \mathrm{\infty }}{lim} \left({\left(\frac{1}{3}\right)}^k+1\right.}\right)\end{array}$

$\begin{array}{r}=\frac{e^{\mathrm{\infty }}}{\left({\left(\frac{1}{3}\right)}^{\mathrm{\infty }}+1\right)}\\ =\frac{0}{(0+1)}\\ =\frac{0}{1}\\ =0\end{array}$