Question

In Problems 51–66, determine whether each infinite geometric series converges or diverges. If it converges, find its sum.

$\sum _{k=1}^{\infty }3{\left(\frac{3}{2}\right)}^{k-1}$

Short answer

The given infinite geometric series $\sum _{k=1}^{\infty }3{\left(\frac{3}{2}\right)}^{k-1}$diverges.

Step-by-step solution

Step 1. Write the given information.

The given geometric series is:

$\sum _{k=1}^{\infty }3{\left(\frac{3}{2}\right)}^{k-1}$

Determine the first term and common ratio.

The first term is$\left(a_1\right)=3$.

The common ratio is the ratio of successive terms:

$r=\frac{3}{2}$

Step 3. Determine whether the infinite geometric series is converges or diverges.

If$\left|r\right|<1$ then the infinite geometric series $\sum _{k=1}^{\infty }{a}_1{r}^{k-1}$converges.

As we can see that $\frac{3}{2}>1$, therefore the given series not converges.

The given infinite geometric series diverges.