# Summation limits change

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?

• 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 – Swapnil Rustagi Dec 24 '16 at 18:22
• @Swapnil, okay thank you! Should I close the question? Otherwise, you can formalize the asnwer and I'll mark it – Euler_Salter Dec 24 '16 at 18:49

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.