A question about geometric series

In a geometric series with first term a, common ratio r, is a=0 allowed? Or is that against the definition of a geometric series? Because everywhere I have searched, the only restriction in the definition of a geometric series is that r=/=0.

Now I say this because in C2, everyone (including the mark scheme) thinks it is ok to simply 'divide both sides of the equation by a", clearly ignoring the fact that they might be dividing by 0. I usually address the case where a=0 and say that if this is true, r can have any value besides 1 (this question is, for example, where they say that the sum to infinity of the series is 4 times the 3rd term). I know you might say that I should 'just assume a=/=0, but I don't consider that proper maths... Help me out?

In a geometric series with first term a, common ratio r, is a=0 allowed? Or is that against the definition of a geometric series? Because everywhere I have searched, the only restriction in the definition of a geometric series is that r=/=0.

Now I say this because in C2, everyone (including the mark scheme) thinks it is ok to simply 'divide both sides of the equation by a", clearly ignoring the fact that they might be dividing by 0. I usually address the case where a=0 and say that if this is true, r can have any value besides 1 (this question is, for example, where they say that the sum to infinity of the series is 4 times the 3rd term). I know you might say that I should 'just assume a=/=0, but I don't consider that proper maths... Help me out?

Think about it. If a=0, what will the next term be... a*r = 0*r = 0

There is no such thing as a geometric sequence with 0 as the first term.

That is the point I'm trying to make.

There is no such thing as a geometric sequence with 0 as the first term.

I know, I thought I would reiterate it

Thats not true.

The sequence: 0,0,0,0,0,...

is still a geometric sequence, so it does exist.

What would "r" be? $\dfrac{0}{0}$? What is this equal to?

Oh okay



What would "r" be? $\dfrac{0}{0}$? What is this equal to?

Not only this but for a geometric series to converge when you sum to infinity by definition |r| < 1. Yet, that series you suggests can have have any value for r, including those greater than |r| yet the sum to infinity will be 0 and not diverge even though r may be greater than 1.

r could be a non zero real number.

Let a=0 so that the sequence is r(0),r^2(0),... and so on.

Would you mind ponting to a past paper/example? There might be something in the question.

I don't consider that to be geometric. The ratio between consecutive terms is defined as:

$\dfrac{u_{n+1}}{u_n}$

So for your series:

$r = \dfrac{k^2 \times 0}{k \times 0} k \in \mathbb{R}$?

I don't consider that to be geometric. The ratio between consecutive terms is defined as:

$\dfrac{u_{n+1}}{u_n}$

So for your series:

$r = \dfrac{k^2 \times 0}{k \times 0} k \in \mathbb{R}$?

Which is an indeterminate form. We wouldn't necessarily know what the value of r is, but that doesn't mean it doesn't exist.
So technically there is still a ratio. We just wouldn't know what it was from the sequence.

Well then there's this argument, which is pretty convincing. Argh!!

That argument is more convincing for saying it is not a geometry series.

Which is an indeterminate form. We wouldn't necessarily know what the value of r is, but that doesn't mean it doesn't exist.
So technically there is still a ratio. We just wouldn't know what it was from the sequence.

So you have an "undefined" ratio say. But in order to define a geometric series, you need to know "a", and "r". If you are given these values you can generate any geometric sequence, these values are what defines the sequence. How would you define the sequence 0, 0, 0 .... ? You would define it differently from every other geometric sequence, because r could be any real number. So is it still geometric?

I understand your reasoning, but I am still unconvinced as you can't get away from the issue of dividing by zero. Interesting discussion though

EDIT: 0, 0, 0 ... is arithmetic in my eyes.