Question

Which geometric series converges?

Answer 1

Answer:

Option (1)

Step-by-step explanation:

In a geometric series,

$\sum_{n=0}^{\infty}ar^n=a+ar^1+ar^2+....$

Here r = $\frac{a_n}{a_{n-1}}$

For |r| < 1, series will converge.

For |r| > 1, series will diverge.

Option (1)

Given geometric series is,

$\frac{1}{2}+\frac{1}{4}+ \frac{1}{8}+ \frac{1}{16}.......$

Common ratio = $\frac{\frac{1}{4}}{\frac{1}{2} }$

                        = $\frac{1}{2}$

Since, $\frac{1}{2}$

Series will converge.

Option (2)

$\frac{1}{2}+1+2+4+......$

r = $\frac{1}{\frac{1}{2}}=2$

Since, 2 > 1,

Series will diverge.

Option (3)

$\frac{1}{2}+ \frac{3}{2}+ \frac{9}{2}+ \frac{27}{2}+......$

Common ratio 'r' = $\frac{\frac{3}{2} }{\frac{1}{2} }=3$

Since, 3 > 1,

Series will diverge.

Option (4)

$\frac{1}{2}+3+18+108+.....$

Common ratio 'r' = $\frac{3}{\frac{1}{2}}=6$

Since, 6 > 1

Series will diverge.

Therefore, Option (1) will be the correct option.