Respuesta :
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.
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 :
[tex]\dfrac{1}{7}+\dfrac{1}{28}+\dfrac{1}{112}+\dfrac{1}{448}+~~.~~.~~.[/tex]
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
[tex]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}.[/tex]
So, we get
[tex]|r|=|\dfrac{1}{4}|=0.25<1.[/tex]
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
[tex]S=\dfrac{a}{1-r}.[/tex]
For the given series,
[tex]a=\dfrac{1}{7},~~r=\dfrac{1}{4}.[/tex]
Therefore, the required sum will be
[tex]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}.[/tex]
Thus, the given series is convergent and it has a sum of [tex]\dfrac{4}{21}.[/tex]
Option (D) is CORRECT.
