Understanding the mathematical relation in the convolution sums

Question

I am looking at an example on convolution sums. In the example it states the following:

$$ y[n] = \sum_{k=-\infty }^{n }2^{k} = \sum_{m=0}^{\infty } \left ( \frac{1}{2} \right )^{m-n}$$

The sum for $2^{k}$ happens for n < 0, however I feel I am missing some mathematical steps in between and have not been able to understand how it leads to $\left ( \frac{1}{2} \right )^{m-n}$ . Any clarification on the operation between these two steps or understanding the relationship is appreciated.

Answer 1

Starting with the right-hand side we obtain \begin{align*} \sum_{m=0}^{\infty}\left(\frac{1}{2}\right)^{m-n} &=\sum_{m=0}^{\infty}2^{n-m}\tag{1}\\ &=\sum_{m=-n}^{\infty}2^{-m}\tag{2}\\ &=\sum_{m=-\infty}^n2^m\tag{3} \end{align*}

Comment:

• In (1) we use $a^{-p}=\left(\frac{1}{a}\right)^p$.

• In (2) we shift the index to start with $m=-n$. To compensate this shift we substitute each occurrence of $m$ with $m+n$ within the scope of the sum.

• In (3) we change the order of summation by substituting $m$ with $-m$.