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$$1+1+1+1+1+...$$

Is it an arithmetic progression or a geometric progression?

My teacher said its neither of them, but a simple sequence. But how I think is that it can be both of them.

It can be an arithmetic progression with initial term=0 and common difference=0

and geometric progression with initial term=0 and common ratio=1

May I know which one is true?

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  • $\begingroup$ If initial term is $0$, then in both cases it becomes $0,0,0, \cdots$. $\endgroup$ – Sahiba Arora Jul 20 '17 at 15:49
  • $\begingroup$ Not an expert on this topic but I would guess that your teacher is right because both expressions imply progression (change). And both of your methods imply that nothing changes. $\endgroup$ – MrYouMath Jul 20 '17 at 15:51
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  • $1+1+1+1+1+...$ is a series;

  • $1,1,1,1,1,\cdots$ is a sequence;

  • the sequence follows an arithmetic progression of initial term $1$ and common difference $0$;

  • the sequence follows a geometric progression of initial term $1$ and common ratio $1$.


Whether a constant series is considered a progression or not is mostly a matter of taste. If not accepted, this will cripple the proofs with conditions $d\ne0$ or $r\ne1$, which is not practical.

If you want to insist that the progressions are non-constant, you can speak of proper progressions.

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Firstly, what you have written is a series, not a sequence (aka progression). A sequence is made up of just the terms in the series, i.e. $(1,1,1,\dotsc)$, without summing them.

To address your actual question, you can indeed view the sum as am arithmetic series with initial term $1$ and difference $0$, or also as a geometric series with initial term $1$ and ratio $1$.

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An arithmetic progression is a sequence $(u_n)_{n \geq 0}$ which is of the form:

$$ \forall n \in \mathbb{N}, \; u_{n+1} = u_n + r $$

where $r \in \mathbb{R}$ and $u_0 \in \mathbb{R}$.

A geometric progression is a sequence $(v_n)_{n \geq 0}$ of the form:

$$ \forall n \in \mathbb{N}, \; v_{n+1} = q v_{n} $$

where $q \in \mathbb{R}$, $v_{0} \in \mathbb{R}$.

If you take $u_0 = 1$ and $r=0$, then you have :

$$ (u_0, u_1, u_2, \ldots ) = (1, 1, 1, \ldots ). $$

If you take $v_0 = 1$ and $q = 1$, you have :

$$ (v_0, v_1, v_2, \ldots) = (1, 1, 1, \ldots). $$

In this case, $(1,1,1, \ldots)$ is both an arithmetic sequence and a geometric sequence.

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Well, if you consider the series "1,1,1.....n" terms to be geometric sequence then like a, ar , .. "a" being initial term and r being the common ratio, then the sum would be equal to
(a×(1-(r^n)))/(1-r), then the sum of 1+1+1....N terms will come to be in zero divided by zero format. So you will have to apply concepts of limits to get the sum.

But if you consider it as arithmetic progression, you will directly get the sum of the series equal to "n".

So it would be more appropriate to considered the series as the arithmetic progression.

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