Geometric series with sum to infinity of 1/t, 1/(t-1), 1/(t+2)

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Homework Statement



Q.: The numbers [itex]\frac{1}{t}[/itex], [itex]\frac{1}{t - 1}[/itex], [itex]\frac{1}{t + 2}[/itex] are the first, second and third terms of a geometric sequence.
Find (i) the value of t,
(ii) the sum to infinity of the series.

Homework Equations



S[itex]\infty[/itex] = [itex]\frac{a}{1 - r}[/itex]

The Attempt at a Solution



I have already solved (i), the value of t = [itex]\frac{1}{4}[/itex].

Ans.: From textbook = 6

Attempt at (ii): S[itex]\infty[/itex] = [itex]\frac{a}{1 - r}[/itex]

a = [itex]\frac{1}{t}[/itex] = [itex]\frac{1}{1/4}[/itex] = 4

r = [itex]\frac{U2}{U1}[/itex] = [itex]\frac{1}{t-1}[/itex]/ 4

[itex]\frac{1}{1/4 - 1}[/itex]/ 4

[itex]\frac{1}{-3/4}[/itex]/ 4

[itex]\frac{-4/3}{4}[/itex]

[itex]\frac{-4}{3}[/itex]([itex]\frac{1}{4}[/itex]) = [itex]\frac{-4}{12}[/itex] = [itex]\frac{-1}{3}[/itex]

Lastly,
S[itex]\infty[/itex] = [itex]\frac{a}{1 - r}[/itex] = [itex]\frac{4}{1-(-1/3)}[/itex]

[itex]\frac{4}{1 + 1/3}[/itex]

[itex]\frac{4}{4/3}[/itex] = 4([itex]\frac{3}{4}[/itex]) = [itex]\frac{12}{4}[/itex] = 3

I have shown this problem on another site, and the other users seem to think that the book has the answer incorrect; with 3 being the correct value. I just wanted to post my attempt here too, to get a second opinion. Thank you.
 
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