Partial Sums and Their Indexing

Why Indexing Matters

A partial sum is a finite sum made from the initial terms of a sequence. The calculation may be straightforward, but its notation can be easy to misread: the upper index of a sum is not always the number of terms it contains. That distinction matters whenever the sequence begins at an index other than one.

For example, the indices \(3,4,5,6,7\) give five terms, not seven. If a partial sum starts at index \(3\) and ends at index \(n\), its number of terms is \(n-3+1\). The added \(1\) accounts for including both endpoints.

This tutorial focuses on how to form a partial sum and determine exactly which terms it contains. It does not address what happens to partial sums as the number of terms grows.

Partial Sums with an Arbitrary Starting Index

Suppose a sequence has terms \(a_k\), defined for integer indices \(k\geq m\). The sum of its terms from index \(m\) through index \(n\) includes both the first term \(a_m\) and the last term \(a_n\). We can denote it by \(S_n\).

The subscript \(n\) in \(S_n\) identifies the final index included. It does not, by itself, state how many terms have been added. The starting index \(m\) is needed to determine that count.

There is another useful way to specify a partial sum: give the number \(N\) of terms to include. Starting with \(a_m\), the first \(N\) terms have indices \(m,m+1,\ldots,m+N-1\). We can write their sum as \(P_N\).

These notations describe the same finite block when their indices correspond. The relation between them is \(P_N=S_{m+N-1}\): a sum of \(N\) terms starting at \(m\) ends at \(m+N-1\). Conversely, if the final index is \(n\), the number of terms is \(N=n-m+1\).

Counting the Terms

To count terms from \(m\) through \(n\), list the indices as an inclusive integer interval. The distance between the endpoints is \(n-m\), but that counts the gaps between indices. Since there is one more index than there are gaps, the number of terms is \(n-m+1\).

For instance, the indices \(4,5,6,7,8,9\) have five gaps between the endpoints and six indices in total. The formula gives \(9-4+1=6\). This endpoint check is often the quickest way to catch an indexing error before doing any arithmetic.

From a Term Count to the Final Index

When a problem asks for the first \(N\) terms, it gives a count rather than a final index. If the first term has index \(m\), move forward \(N-1\) index steps to reach the last term. Therefore, the final index is \(m+N-1\). The subtraction of \(1\) is necessary because the first included term is already present before any steps are taken.

From a Final Index to a Term Count

Some problems give the upper index instead of the number of terms. If the sum starts at \(m\) and ends at \(n\), count inclusively by calculating \(n-m+1\). This count remains valid whether the starting index is zero, one, or another integer.

Why the Index Formulas Work

The formulas for counting terms and finding the endpoint follow directly from the way consecutive indices are arranged. Here is a derivation that also explains why the endpoint formulas are inverses of one another.

This reasoning establishes both directions: a specified count determines the final index, and specified endpoints determine the count. It also shows why neither formula depends on the values of the terms. The indexing describes which terms are included; the formula for \(a_k\) determines what those terms equal.

If the terms are written explicitly rather than with a formula, the same indexing logic applies. For example, a list beginning at \(a_4\) and containing five terms is \(a_4,a_5,a_6,a_7,a_8\). The last index is \(4+5-1=8\). Whether the terms are numbers, expressions, or values from a sequence, the count is determined by the indices.

Re-indexing the Same Block

A finite block can be described using a different index variable, provided every original index is matched to exactly one new index. For a sum starting at \(m\), set \(k=m+j\). When \(k=m\), \(j=0\); when \(k=m+N-1\), \(j=N-1\). Thus the same terms can be written with \(j\) running from \(0\) to \(N-1\):

This equality is a change of labels, not a change in the terms being added. The lower and upper limits must both be adjusted according to \(k=m+j\).

A Common Indexing Mistake

A frequent error is to treat the upper index as the number of terms. For example, a student might claim that the sum from \(k=2\) to \(k=8\) contains six terms, because \(8-2=6\). That calculation counts the gaps between the endpoints, not the included indices.

The incorrect count can cause the last term to be dropped. For instance, if \(a_k=k\), the wrong calculation might add only \(2+3+4+5+6+7=27\). But the requested sum includes \(a_8=8\) as well:

Check: the indices \(2,3,4,5,6,7,8\) number seven, as \(8-2+1=7\) predicts. The incorrect answer \(27\) omitted the endpoint term \(8\); adding it gives \(27+8=35\). Always check both the first and last included indices before evaluating a sum.

Keeping the Two Conventions Straight

There are two common ways to describe a partial sum: by its final index or by its number of terms. Neither convention is difficult once the starting index is stated. The potential confusion arises when the same letter or subscript is assumed to mean both things.

For a sequence beginning at index \(1\), the sum from \(1\) through \(n\) contains \(n\) terms. That familiar special case can make it tempting to identify an upper index with a term count in every situation. When the starting index is not \(1\), use the full count \(n-m+1\) instead.

Summary

To form a partial sum reliably, first identify its starting index and determine whether the problem gives a final index or a number of terms. If it gives endpoints \(m\) and \(n\), the count is \(n-m+1\). If it gives \(N\) terms starting at \(m\), the final index is \(m+N-1\). Then write or evaluate exactly the terms in that inclusive index range.