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I know this is a very naive question, but I still struggle with summation limits.

How does someone go from the first summation to the second? I am looking at the second equation, second step. We want to make the index start at $0$ instead of $-m$, so we create a new variable $j = n+m$ so that when $n = -m$ we get $j=0$. However this doesn't seem to work as substituting we would get: $$\sum_{j=0}^{\infty}c_{j-m}(z-a)^{j}$$ is it then $n$ a dummy index in the last step? And is my method correct?

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  • $\begingroup$ Yes, your method is correct. And absolutely, n is a dummy index here. Actually, the index used with the sigma symbol is always a dummy index. If you try writing a few terms of this sum, you will immediately see it $\endgroup$ – Swapnil Rustagi Dec 24 '16 at 18:22
  • $\begingroup$ @Swapnil, okay thank you! Should I close the question? Otherwise, you can formalize the asnwer and I'll mark it $\endgroup$ – Euler_Salter Dec 24 '16 at 18:49
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Yes, your method is correct.

And absolutely, $n$ is a dummy index here. Actually, the index used with the sigma symbol is always a dummy index. If you try writing a few terms of this sum, you will immediately see it.

/* Posted as answer in response to comment by OP. If this is inappropriate, please do not downvotes, ask for deletion in comments*/

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