In my notes I saw this equation and I can only speculate how it works:
$$\sum^\infty_{n=1}\left(\frac{13}{12}\right)^{-(n+2)} = \sum^\infty_{k=3}\left(\frac{13}{12}\right)^{-k} $$
The following step $$\sum^\infty_{k=3}\left(\frac{13}{12}\right)^{-k} = \sum^\infty_{k=3} \left(\frac{12}{13}\right)^k$$ is clear to me but I dont understand the equation above.
Can someone clear this up for me ?
Note that both sides of the first equality unfold to $$ \Big(\frac{13}{12}\Big)^{-3} + \Big(\frac{13}{12}\Big)^{-4} + \Big(\frac{13}{12}\Big)^{-5} + \Big(\frac{13}{12}\Big)^{-6} + \cdots $$ The terms are just indexed differently: $$ \underbrace{\Big(\frac{13}{12}\Big)^{-3}}_{n=1} + \underbrace{\Big(\frac{13}{12}\Big)^{-4}}_{n=2} + \underbrace{\Big(\frac{13}{12}\Big)^{-5}}_{n=3} + \underbrace{\Big(\frac{13}{12}\Big)^{-6}}_{n=4} + \cdots $$ versus $$ \underbrace{\Big(\frac{13}{12}\Big)^{-3}}_{k=3} + \underbrace{\Big(\frac{13}{12}\Big)^{-4}}_{k=4} + \underbrace{\Big(\frac{13}{12}\Big)^{-5}}_{k=5} + \underbrace{\Big(\frac{13}{12}\Big)^{-6}}_{k=6} + \cdots $$