Is it possible to approximate the following series in a similar way to the Neumann series for example where the formula is the limit of the series as the terms approach infinity: $$ \sum_{i=0}^{\infty} \sum_{j=0}^{\infty}a^ib^j $$ $a$ and $b$ are both fractions.
Asked
Active
Viewed 56 times
3 Answers
2
Try $\displaystyle\left(\sum_{i=0}^{\infty}a^i\right)\left( \sum_{j=0}^{\infty}b^j\right)$ as the product of geometric series if $-1 \lt a \lt 1$ and $-1 \lt b \lt 1$.
You should not need an approximation.
Henry
- 157,058
2
If I am not wrong, this is simply a geometric series. \begin{align} \sum_{i \ge 0} \sum_{j \ge 0}a^i b^j &=\sum_{i \ge 0} a^i \sum_{j \ge 0}b^j\\ &= \sum_{i \ge 0} \frac{a^i}{1-b}\\ &=\frac{1}{(1-a)(1-b)} \end{align}
SuperAbound
- 5,554
1
$$ \sum_{i=0}^\infty \left( \underbrace{\sum_{j=0}^\infty a^ib^j}_\text{Let's work with this inside part first.} \right) $$ $$ \sum_{j=0}^\infty ( a^ib^j) $$ As $j$ goes from $0$ to $\infty$, $i$ does not change, so $a^i$ does not change. Therefore this sum is $$ a^i \sum_{j=0}^\infty b^j. $$ Now we have $$ \sum_{i=0}^\infty \left( a^i \sum_{j=0}^\infty b^j \right). $$ As $i$ goes from $0$ to $\infty$, the factor $\sum_{j=0}^\infty b^j$ does not change. Therefore it can be pulled out of the outer sum in the same way.
