Thread: Sum to Infinity of a Geometric Series

1. Sum to Infinity of a Geometric Series

Q.: The numbers 1/ t, 1/ (t - 1), 1/ (t + 2) 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.

Eq.: S¥ = a/ 1 - r

Ans.: (i) t = 1/ 4, (ii) S¥ = 6

Attempt:
(i) U2/ U1 = U3/ U2
(1/ (t - 1))/ (1/ t) = (1/ (t + 2))/ (1/ (t - 1))
(1/ (t - 1)) (1/ (t - 1)) = (1/ t) (1/ (t + 2))
(1/ (t^2 - 2t + 1)) = (1/ (t^2 + 2t))
(1/ (t^2 - 2t + 1 - t^2 - 2t))
1/ (-4t + 1) … this is where I lose the flow, slightly. Can someone help me confirm if my next move is permissible?

1/ (4t + 1)
1 = 4t
t = 1/ 4

(ii) S¥ = a/ 1 - r
a = 1/ (1/ 4) = 4, i.e. 1/ t
r = U2/ U1 or U3/ U2 = (1/ ((1/ 4) - 1))/ 4 = (1/ (-3/ 4))/ 4 = (-4/ 3)/ 4 =
(-4/ 3)(1/ 4) = -4/ 12 = -1/ 3
4/ (1 - (-1/ 3)) … This is where I think I may have gone wrong.
4/ (1 + (1/ 3)) … I think I need to have 1 - (1/ 3) here but I can’t see how to bring that about.

4/ (4/ 3)
4 (3/ 4)
12/ 4 = 3 The answer here should be 6 but I can only get 3.

If I can get 1 - (1/ 3), above, I’ll have (4/ (2/ 3)), and that gives 6 but as I have shown, the question does not work out that way for me. Have I missed something? Thank you.

2. Re: Sum to Infinity of a Geometric Series

I disagree that the sum is 6.

$\frac{1}{t} , \frac{1}{t-1} , \frac{1}{t+2} , ...$

common ratio of the first two terms ... $r = \frac{t}{t-1}$

common ratio of the 2nd and 3rd terms ... $r = \frac{t-1}{t+2}$

$\frac{t}{t-1} = \frac{t-1}{t+2}$

$(t-1)^2 = t(t+2)$

$t^2 - 2t + 1 = t^2 + 2t$

$1 = 4t$

$t = \frac{1}{4}$

$r = -\frac{1}{3}$

infinite sum ...

$\frac{4}{1 - \left(-\frac{1}{3}\right)} = 3$

3. Re: Sum to Infinity of a Geometric Series

$t=\frac{1}{4}$ is correct, so the geometric sequence becomes: $4,\frac{-4}{3}, \frac{4}{9},...$
So $r=\frac{-1}{3}$ and because $\left|\frac{-1}{3}\right|<1$ we can use: $S=\frac{a}{1-r}=\frac{4}{1+\frac{1}{3}}=3$

So you're correct!

4. Re: Sum to Infinity of a Geometric Series

Wow! I'm really surprised that they got that one wrong, in the book. Thanks for all the help, guys.