Partial sums.

Some sequences naturally arise as the sum of terms of another sequence.

Example 2.1.5.

Sam keeps track of how many push-ups she does each day of her “do lots of push-ups challenge.” Let \((a_n)_{n \ge 1}\) be the sequence that describes the number of push-ups done on the \(n\)th day of the challenge. The sequence starts

\begin{equation*} 3, 5, 6, 10, 9, 0, 12, \ldots\text{.} \end{equation*}

Describe a sequence \((b_n)_{n \ge 1}\) that describes the total number of push-ups done by Sam after the \(n\)th day.

Solution.

Given any sequence \((a_n)_{n \in \N}\text{,}\) we can always form a new sequence \((b_n)_{n \in \N}\) by

\begin{equation*} b_n = a_0 + a_1 + a_2 + \cdots + a_n\text{.} \end{equation*}

Since the terms of \((b_n)\) are the sums of the initial part of the sequence \((a_n)\) ways call \((b_n)\) the sequence of partial sums of \((a_n)\). Soon we will see that it is sometimes possible to find a closed formula for \((b_n)\) from the closed formula for \((a_n)\text{.}\)

To simplify writing out these sums, we will often use notation like \(\d\sum_{k=1}^n a_k\text{.}\) This means add up the \(a_k\)'s where \(k\) changes from 1 to \(n\text{.}\)

Example 2.1.6.

Use \(\sum\) notation to rewrite the sums:

  1. \(\displaystyle 1 + 2 + 3 + 4 + \cdots + 100\)

  2. \(\displaystyle 1 + 2 + 4 + 8 + \cdots + 2^{50}\)

  3. \(6 + 10 + 14 + \cdots + (4n - 2)\text{.}\)

Solution.

If we want to multiply the \(a_k\) instead, we could write \(\d\prod_{k=1}^n a_k\text{.}\) For example, \(\d\prod_{k=1}^n k = n!\text{.}\)

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