Respuesta :

irspow
The series will converge if r^2<1, where r is the common ratio of the geometric sequence.

The first has an r=2, so it diverges

The second has an r=-4 so it diverges.

The third has an r=3 so it diverges.

The fourth has an r=1/2 so it converges.

So only the last one converges...

The geometric series 1 + [tex]\frac{1}{2} + \frac{1}{4} +\frac{1}{8}[/tex] +...... converging geometric series.

What is geometric series?

A geometric series is "the sum of the infinite number of terms that have common ratio between successive terms".

According to the question,

Geometric series  ∑a rⁿ⁻¹  where 'a' is the coefficient and 'r' is the common ratio.

The converging geometric series depends on the value of 'r' if |r| < 1, then the geometric series converges. If |r|>1, then the geometric series diverges. If the |r| = 1, the geometric series diverges. If the |r| = -1, then the geometric series oscillates between 1 and -1.

  • The series ∑ [tex]\frac{1}{5}2^(n-1)[/tex] has the common ratio 2 > 1, then the geometric series diverges.
  • The series ∑7(-4)0ⁿ⁻1 has the common ratio -4 < 1, then the geometric series diverges.
  • The series [tex]\frac{1}{81} + \frac{1}{27} + \frac{1}{9} +\frac{1}{3}+ ......[/tex] has the common ratio 1/27÷1/81 = 3, the common ratio 3 > 1, then the geometric series diverges.
  • The series [tex]\frac{1} + \frac{1}{2} + \frac{1}{4} +\frac{1}{8}+ ......[/tex] has the common ratio 1/2÷1 = 1/2, the common ratio 1/2 < 1, then the geometric series diverges.

  Hence, the series [tex]\frac{1} + \frac{1}{2} + \frac{1}{4} +\frac{1}{8}+ ......[/tex] has the common ratio 1/2÷1 = 1/2, the common ratio 1/2 < 1, then the geometric series diverges.

Learn more about geometric series here

https://brainly.com/question/12980130

#SPJ2