Question

Which geometric series converges? StartFraction 1 Over 81 EndFraction StartFraction 1 Over 27 EndFraction one-ninth one-third ellipsis 1 one-half one-fourth one-eighth ellipsis Sigma-Summation Underscript n = 1 Overscript infinity EndScripts 7 (negative 4) Superscript n minus 1 Sigma-Summation Underscript n = 1 Overscript infinity EndScripts one-fifth (2) Superscript n minus 1.

Answer 1

The geometric series that converges from the listed option is $1+\frac{1}{2} + \frac{1}{4} +\frac{1}{8} ,...$ Option B is correct.

The nth term of a geometric sequence is expressed as;

$a_n=ar^{n-1}$

• a is the first term of the sequence

• r is the common ratio

• n is the number of terms.

For a geometric sequence to converge, the modulus of its common ratio must be less than 1 (|r| < 1), otherwise, it diverges.

For the first series given as $\frac{1}{81} + \frac{1}{27} +\frac{1}{9} +\frac{1}{3}+ ,...$

$r = \frac{1}{27} \div \frac{1}{81}\\r=\frac{1}{27} \times 81\\r=\frac{81}{27} \\r=3 > 1$

Since the common ratio of the sequence is greater than 1, hence the series diverges.

For the series $1+\frac{1}{2} + \frac{1}{4} +\frac{1}{8} ,...$

$r=\frac{1/2}{1} =\frac{1/4}{1/2} = \frac{1}{2}$

Since the common ratio of the sequence is less than 1, hence the series converges.

Learn more on convergence of series here: brainly.com/question/14294471