are $\sum_{i=0}^{n}i$ and $\sum_{i=n}^{0}i$ equivalent?

Question

So here's the ugly history of how I came to ask this question. I was following this proof:

and got stuck at this step:

$$\sum_{j=0}^{(\log_2n) - 1}\frac{1}{(\log_2n) - j} = \sum_{l = 1}^{\log_2n}\frac{1}{l}$$

I tried a simple change in variables, but was confused because the bounds didn't match.

We know that the step holds true because another student did a proof that makes sense.

But now my question is are $\sum_{i=0}^{n}i$ and $\sum_{i=n}^{0}i$ equivalent? If that's true, I could've saved myself a lot of trouble.

Answer 1

Yes, they are. It is rare to see the indices written in that order, but you have all the same terms in each sum.