Question

23)   Does the following infinite geometric series diverge or converge? Explain.

Answer 1

Answer:  The correct option is (D) It converges; it has a sum.

Step-by-step explanation:  We are given to check whether the following infinite geometric series diverge or converge :

$\dfrac{1}{7}+\dfrac{1}{28}+\dfrac{1}{112}+\dfrac{1}{448}+~~.~~.~~.$

We know that

an infinite geometric series converges if the modulus of the common ratio is less than 1.

For the given geometric series, the common ratio r is given by

$r=\dfrac{\frac{1}{28}}{\frac{1}{4}}=\dfrac{\frac{1}{112}}{\frac{1}{28}}=\dfrac{\frac{1}{448}}{\frac{1}{112}}=~~.~~.~~.~~=\dfrac{1}{4}.$

So, we get

$|r|=|\dfrac{1}{4}|=0.25$

Therefore, the given infinite geometric series converges.

Also, we know that the sum of a convergent infinite geometric series with first term a and common ratio r is given by

$S=\dfrac{a}{1-r}.$

For the given series,

$a=\dfrac{1}{7},~~r=\dfrac{1}{4}.$

Therefore, the required sum will be

$S=\dfrac{a}{1-r}=\dfrac{\frac{1}{7}}{1-\frac{1}{4}}=\dfrac{\frac{1}{7}}{\frac{3}{4}}=\dfrac{1}{7}\times\dfrac{4}{3}=\dfrac{4}{21}.$

Thus, the given series is convergent and it has a sum of $\dfrac{4}{21}.$

Option (D) is CORRECT.

Answer 2

It diverges; it does not have a sum.  the sum of a geometric series is given by a /(1-r) where a is the first term and the r is the ratio between the terms. r=3 in the series you have given . The absolute value of r must be less then 1 for a geometric series to converge.