Geometric series C2

Question:
All the terms of a certain geometric series are positive. The first term is a and the second term is $a^{2}-a$. Find the set of values of a for which the series converges.


AEB NOV.87 (PAPER)

My answer was 0<a<1
but the book answer is 1<a<2 why? I thought that since all the series are positive then it is 0<a<1 but i must be wrong i want to understand the reasoning behind the answer given by the book


Thank you very much for helping

When does a geometric series converge?

What is the value you need to calculate and use here?

Question:
All the terms of a certain geometric series are positive. The first term is a and the second term is a^{2}-a . Find the set of values of a for which the series converges.

Given that a= $\frac{5}{3}$

a) find the sum of the first 10 terms of the series, giving your answers to 2 decimal places
b)show that the sum to infinity of the series is 5
c) find the least number of terms of the series required to make their sum exceed 4.999

that's the full question

Okay. I'll ask the same questions as I did in my previous post.

If I had a series a, ar.... What is the condition for the series converging? How will that help to answer the question?

the condition is that r being the common ratio is -1<r<1 in order for the series to converge
the series will converge when it tends towards a specific value
when I sub a = 5/3 into the sum to infinity formula i get 5/9


To be honest im a bit confused. All i know is that convergence means that there is a sum to infinity.

I think i have the answer please bare with me

I decided to do: $r= \frac{U_{2}}{U_{1}}= \frac{a^{2}-a}{a}=a-1$
$r=a-1$

if the series is to converge (remembering that GP is positive) thus
0<r<1

thus:
$0<r<1=a-1$

$(0<r<1)+1=a[br]\therefore 1<r<2=a [br]\therefore 1<a<2$

thats my answer or my thoughts

Yes, well done

A hint you could have seen is that they give 5/3 as r later, so your range would have to contain that.

And if you quote users when replying it saves them time and possibly not responding as they don't know that you have responded.

thank you very much I never thought anyone will be up at this time

one question

shouldn't this be:

$(0<r<1)+1=a[br]\therefore 1<r+1<2=a [br]\therefore 1<a<2$


instead of


$(0<r<1)+1=a[br]\therefore 1<r<2=a [br]\therefore 1<a<2$

Perhaps I'm being dense but this notation looks gibberish; (0<r<1) does not represent a number.

I'd say:



$0 < a-1 < 1[br]\therefore 1<a<2$