Section 9.3 Manipulation of Series
This short section covers a couple useful algebraic manipulations of sigma notation and infinite series. These algebraic tools are useful for understanding series, determining convergence, and combining different series together.
First, series are linear as long as the indices match up. Series split up over sum and differences in the term. Constants present in the term can be factored out of the sigma notation.
\begin{align*}
c \sum_{n=0}^\infty a_n \amp = \sum_{n=0}^\infty ca_n\\
\sum_{n=0}^\infty (a_n \pm b_n) \amp = \sum_{n=0}^\infty a_n
\pm \sum_{n=0}^\infty b_n
\end{align*}
Second, since sigma notation is just notation for a sum, pieces of the sum can be written explicitly using the ordinary notation for sums. This is called removing terms from the series.
\begin{equation*}
\sum_{n=0}^\infty a_n = a_0 + a_1 + a_2 + \sum_{n=3}^\infty a_n
\end{equation*}
Third, I can shift the indices. The key idea here is balance: whatever I do to the index in the bounds, I must do the opposite to the index in the terms to balance it out.
\begin{equation*}
\sum_{n=0}^\infty a_n = \sum_{n=1}^\infty a_{n-1} =
\sum_{n=-1}^\infty a_{n+1}
\end{equation*}
Note that the linearity properties required the bounds of the series to match up. Shifting and/or removing terms is very useful for adjusting series so that they have the same starting bounds, and thus can be combined into a single series.
Example 9.3.1.
In this example, I want to add two series which don’t have matching indices. I shift the first series to make the indices match and allow the addition. In the last step below, I also show an example of common denominator with factorials, which is another frequently useful bit of algebra. In the second term, I need to multiply by all the missing term to make the factorial in the denominator.
\begin{align*}
\sum_{n=3}^\infty \frac{3^n}{n!} + \sum_{n=1}^\infty
\frac{1}{n(n+2)} \amp = \sum_{n=1}^\infty
\frac{3^{n+2}}{(n+2)!} + \sum_{n=1}^\infty
\frac{1}{n(n+2)}\\
\amp = \sum_{n=1}^\infty \left( \frac{3^{n+2}}{(n+2)!} +
\frac{1}{n(n+2)} \right)\\
\amp = \sum_{n=1}^\infty \left( \frac{3^{n+2} +
(n-1)!(n+1)}{(n+2)!} \right)
\end{align*}
