Suppose I have $$\sum_{k=n_1}^{n_2} a_k$$ if I define $$k=n-k_1$$ I can write $$\sum_{k_1=n-n_1}^{n-n_2} a_{n-k_1}$$ Then to better assure myself that I understood index shift correctly I write the following system:
$$k=n-k_1$$ $$k\geq{n_1}$$ $$k\leq{n_2}$$ It can be resolved to: $$k=n-k_1$$ $$k_1\leq{n}-{n_1}$$ $$k_1\geq{n}-{n_2}$$
What is confusing for me here is that inequality signs should be inverted(at least how I understand that) since $k_1$ is in range $n-n_1...n-n_2$. So it seems to me that I didn't understand something correctly although index shift is done formally correct here. Could you please point out my mistake?
Since $n_{1}\leq n_{2}$ it holds that $n-n_{2}\leq n-n_{1}$ so the correct range for $k_1$ is $$n-n_{2}\leq k_{1} \leq n -n_{1}.$$ However, what you have written in the sum is still correct because it preserves the order in which the terms are summed up. Since $n-k_{1}$ is smaller for larger values of $k_{1}$, it follows that, as we go from the larger $n-n_{1}$ to the smaller $n-n_{2}$ we are in fact increasing the index of $a$ at each term.