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A very basic question, but I am missing something...

Consider the series $\sum_{i=1}^{n}a_{n-i} = a_{n-1}+a_{n-2}+\dots+a_0$, which may therefore equivalently written $\sum_{i=0}^{n-1}a_i$.

How does one arrive at the equivalence between the two summation representations by reindexing/change of variables? If we let $k=n-i$, then the lower limit of the sum at $i=1$ corresponds to $k=n-1$ and the upper limit at $i=n$ corresponds to $k=0$, which gives us an empty series $\sum_{k=n-1}^0a_k$.

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You got it almost right. You wrote:

"If we let $k=n-i$, then the lower limit of the sum at $i=1$ corresponds to $k=n-1$ and the upper limit at $i=n$ corresponds to $k=0$"

and which changes the order of summation since rising $i$ corresponds to falling $k$.

This means you have to compute the new sum limits (which you did) and to exchange upper and lower limit,

which gives the series $\sum_{k=0}^{n-1}a_k$.

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  • $\begingroup$ Is it always the case that, if the change of variable from $i$ to $f(i)$ is monotonically decreasing, then the upper and lower limits must be exchanged? I could find a reference for this fact, even though it might be obvious. $\endgroup$ – jII Feb 10 '17 at 22:02
  • $\begingroup$ Yes, it's the case. I think a formal reference would be hard to find since it's pretty clear. $\endgroup$ – Andreas Feb 12 '17 at 9:54

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