How to change upper and lower limits of summation

Question

Given a problem like this one $$\displaystyle \sum_{i=-20}^0 \left(\dfrac{1}{3}\right)^i$$ what would I have to do to the summation to make it go from 0 to 20? I'm assuming you can't just switch the limits without doing anything else to the summation.

Answer 1

Define $n = -i$. We transform the lower and upper bounds and the summation appropriately, expressing them in terms of $n$:

• $i = -20$ becomes $n = 20$
• $i=0$ becomes $n = 0$
• $(1/3)^i$ becomes $(1/3)^{-n}$. Equivalently, $3^n$.

Since summing in the usual sense is commutative, we can also swap the upper and lower bounds without issue and thus claim:

$$\sum_{i=-20}^0 \left( \frac{1}{3} \right)^i = \sum_{n=0}^{20} 3^n$$

Answer 2

Sometimes it might help to write bounds as inequality and do the substitution step by step: $$ \sum_{i=-20}^0 (1/3)^i=\sum_{-20 \leq i \leq 0} (1/3)^i\underset{i:=-t}{=}\sum_{-20 \leq -t \leq 0} (1/3)^{-t}=\sum_{0 \leq t \leq 20} (1/3)^{-t}=\sum_{t=0}^{20} 3^t $$ This is helpful also for index shifting.