Hi all -

I realize this problem might just look like something from precalculus --- but because I'm not asking about the actual solution but rather the "analysis" of one part of it --- I thought to post here. I'm pretty comfortable with my own solution.

But my own work oesn't actually tell/explain/show me why the hint $\displaystyle \left( \text{ original r } = \text{ (new r) } - 1 )\right$ actually works. Can someone please explain how I could come up/infer/guess/calculate the hint by myself --- if the question didn't give it to me? All I can understand is that the problem is trying to use the result in the second expression to prove the last expression. Thank you---

My work which looks right:

$\displaystyle \text{Substituting }r={{r}_{0}}-1\text{ into second expression gives } \\$

$\displaystyle \sum\limits_{{{r}_{0}}-1=1}^{\infty }{\arctan \left( \frac{1}{{{\left( {{r}_{0}}-1 \right)}^{2}}+({{r}_{0}}-1)+1} \right)}=\sum\limits_{{{r}_{0}}=2}^{\infty }{\arctan \left( \frac{1}{{{r}_{0}}^{2}-{{r}_{0}}+1} \right)}$

$\displaystyle \text{So }\sum\limits_{{{r}_{1}}=1}^{\infty }{\arctan \left( \frac{1}{{{r}_{0}}^{2}-{{r}_{0}}+1} \right)}=\arctan \left( \frac{1}{0-0+1} \right)+\sum\limits_{r=2}^{\infty }{\arctan \left( \frac{1}{{{r}^{2}}-r+1} \right)} $

$\displaystyle =\frac{\pi }{4}+\frac{\pi }{4}\text{ so this works perfectly}\text{.} $

Actual question ---