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Suppose I want to show

$$\sum_{n=1}^{\infty}ar^{n-1} = \sum_{n=0}^{\infty}ar^{n}$$

Is it acceptable to increment $n$ by redefining it in terms of itself, i.e., $n = n+1$ (written n+=1 in many programming languages), and thus write

$$ \sum_{n=1}^{\infty}ar^{n-1} = \sum_{n+1=1}^{\infty}ar^{(n+1)-1} = \sum_{n=0}^{\infty}ar^{n}$$

Or is it better to define a dummy variable $m=n-1 \implies n=m+1$ and proceed as follows?

$$\sum_{n=1}^{\infty}ar^{n-1}=\sum_{m+1=1}^{\infty}ar^{(m+1)-1}=\sum_{m=0}^{\infty}ar^{m}=\sum_{n=0}^{\infty}ar^{n}$$

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    $\begingroup$ To do this formally there would probably be two steps: first a substitution $m=n-1$ on the sum and second renaming the bound variable $m$ to $n$ (en.wikipedia.org/wiki/Free_variables_and_bound_variables). However it's very common to do both together as in your first alternative (just be careful not to confuse the "new $n$" with the "old $n$"). $\endgroup$
    – stewbasic
    Commented May 30, 2017 at 22:58
  • $\begingroup$ @stewbasic As justification for the last step I was thinking that at each iteration $x$, the notation defines $n=x$ and $m=x$ and thus by transitivity $\forall x(n=m)$.... $\endgroup$
    – Evan Rosica
    Commented May 30, 2017 at 23:01
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    $\begingroup$ It is also common to use arrows instead of the equal sign, like "$n\leftarrow n+1$", as the equal sign isn't an assignment sign in mathematics. $\endgroup$
    – user251257
    Commented May 30, 2017 at 23:03
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    $\begingroup$ @EvanRosica yes, but it is often used to declare a new variable. In computer science it is also used as assignment. $\endgroup$
    – user251257
    Commented May 30, 2017 at 23:06
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    $\begingroup$ Easier would be to go: $\sum_{n=1}^{\infty}ar^{n-1} = \sum_{n-1=0}^{\infty}ar^{n-1} = \sum_{n=0}^{\infty}ar^{n}$ $\endgroup$
    – farruhota
    Commented Jun 1, 2017 at 11:09

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