Write out the first eight terms of the series to show how the series starts. Then find the sum of the series or show that it diverges.

Summation from n equals 1 to infinity left parenthesis 2 minus StartFraction 3 Over 4 Superscript n EndFraction right parenthesis

The first right terms are

5/4, 13/8, 7/4, 29/16, 37/20, 15/8, 53/28, 61/32.

And the series is divergent.

Step-by-step explanation:

Given the series

S = ∑ (2 - 3/4n) from n = 1 to infinity

For n = 1

S1 = 2 - 3/4 = 5/4

S2 = 2 - 3/8 = 13/8

S3 = 2 - 3/12 = 7/4

S4 = 2 - 3/16 = 29/16

S5 = 2 - 3/20 = 37/20

S6 = 2 - 3/24 = 15/8

S7 = 2 - 3/28 = 53/28

S8 = 2 - 3/32 = 61/32

So, the first 8 terms are:

5/4, 13/8, 7/4, 29/16, 37/20, 15/8, 53/28, 61/32.

Let us show that it diverges.

Let R = lim (a_n) as n - --> infinity

a_n = 2 - 3/4n

R = lim (2 - 3/4n) as n - -> infinity.

R = 2 - 0 = 2

Since R ≠ 0, we conclude that the series is divergent.