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    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?
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    (Original post by PrimeLime)
    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
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    (Original post by PrimeLime)
    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?
    (Original post by lizard54142)
    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.
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    (Original post by Jai Sandhu)
    There is no such thing as a geometric sequence with 0 as the first term.
    That is the point I'm trying to make.
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    (Original post by lizard54142)
    That is the point I'm trying to make.
    I know, I thought I would reiterate it
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    (Original post by Jai Sandhu)
    There is no such thing as a geometric sequence with 0 as the first term.
    Thats not true.

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

    is still a geometric sequence, so it does exist.
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    (Original post by Jai Sandhu)
    I know, I thought I would reiterate it
    Oh okay

    (Original post by Xin Xang)
    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?
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    (Original post by lizard54142)
    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.
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    (Original post by lizard54142)
    Oh okay



    What would "r" be? \dfrac{0}{0}? What is this equal to?
    r could be a non zero real number.

    Let a=0 so that the sequence is r(0),r^2(0),... and so on.
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    Would you mind ponting to a past paper/example? There might be something in the question.
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    (Original post by Jai Sandhu)
    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.
    I may be mistaken but I read that condition as, "if you want the sum to infinity to converge, the magnitude of r must be less than 1". But if the series is already going to converge, surely that condition is not necessary?
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    (Original post by Xin Xang)
    r could be a non zero real number.

    Let a=0 so that the sequence is r(0),r^2(0),... and so on.
    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}?
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    (Original post by SeanFM)
    Would you mind ponting to a past paper/example? There might be something in the question.
    Absolutely nothing in the question or even the specification for that matter (yes, I went as far as to check the specification for a note about the a=0 case).
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    (Original post by lizard54142)
    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.
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    (Original post by lizard54142)
    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}?
    Whilst Xin Xang gave the same reasoning as I would have given, looking at the rigorous definition of the common ratio does seem to shed light on this problem. I was confused why in the formal definition of a convergent infinite geometric series, it was stated that |r|<1 is an 'if and only if' condition and so is always the case for all such series. I guess a=0 would be a contradiction to this statement and I see now that it must be because of the way the common ratio is formally defined. It would lead to an 'indeterminate' common ratio rather than (the subtly different) statement the common ratio can be any number (besides 1, of course).
    But I still don't know why they don't specify a=/=0 in the formal definition.
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    (Original post by Xin Xang)
    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!!
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    (Original post by PrimeLime)
    Well then there's this argument, which is pretty convincing. Argh!!
    That argument is more convincing for saying it is not a geometry series.
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    (Original post by Jai Sandhu)
    That argument is more convincing for saying it is not a geometry series.
    Why is that?
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    (Original post by Xin Xang)
    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.
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    (Original post by lizard54142)
    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.
    The problem is that I can't find any (hyper)formal definitions of geometric series...
 
 
 
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