I'd like to know a method to do a change of variables for sumations; as far as I am concerned it should be analogous to change of variables when integrating a continuous function. I got this summation: \begin{equation} \sum_{k=-\infty}^n 2^k\left(\frac{1}{4}\right)^{n-k} =\sum_{r=0}^{\infty} 2^{n-r}\left(\frac{1}{4}\right)^{r} \end{equation} I started with the left side and got to the right one by mere intuition but when writing down my relations $r=n-k$, I just can't find a way to do it algebraically; I suspect I'm missing some important property but this is like the $3^{\text{rd}}$ time I've try to understand this, so I'd really appreciate some extra help. Thanks.
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1 Answer
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By $n-k=r$ we obtain
$$\sum_{k=-N}^n 2^k\left(\frac{1}{4}\right)^{n-k} = \sum_{r=n+N}^0 2^{n-r}\left(\frac{1}{4}\right)^{r} =\sum_{r=0}^{n+N} 2^{n-r}\left(\frac{1}{4}\right)^{r}$$
and taking the limit as $N\to \infty$ we obtain the identity.
