Index Shifts in Geometric Series

Why Index Shifts Matter

A geometric sum can be written with many different starting indices. For instance, a sum might begin at \(k=3\), while the exponent of its common ratio begins at \(k-1\). The notation may look unfamiliar, but the underlying pattern is still a first term followed by repeated multiplication by the same ratio.

The essential question is not whether the first index is zero or one. It is: what is the actual first term, and what common ratio generates the later terms? Identifying these correctly lets us rewrite the sum in a standard form without adding or removing any terms.

Key idea: When an index is shifted, calculate the term at the original starting index. That term is the new first term; the common ratio stays the same.

Reindexing a Finite Geometric Sum

Suppose \(m\) and \(n\) are integers with \(m\le n\), and consider a geometric sum whose terms have the form \(c r^{k-p}\), where the displayed powers are defined. Its first term occurs at \(k=m\), so it is \(c r^{m-p}\). Set \(j=k-m\). Then \(k=m\) corresponds to \(j=0\), and \(k=n\) corresponds to \(j=n-m\).

\sum_{k=m}^{n} c r^{k-p} = c r^{m-p}\sum_{j=0}^{n-m} r^j.

This is a change of index, not a change to the terms being added. The new sum has \(n-m+1\) terms, just as the original sum does. In the standard first-term form, if the first term is \(A\) and the common ratio is \(r\), the same identity is

\sum_{j=0}^{N-1} A r^j = A\sum_{j=0}^{N-1}r^j, \qquad N=n-m+1.

For \(r\ne1\), the finite geometric-sum formula gives the value. If \(r=1\), every term equals \(A\), so the sum is \(NA\).

\sum_{j=0}^{N-1} A r^j = \begin{cases} A\dfrac{1-r^N}{1-r}, & r\ne 1,\\[6pt] NA, & r=1. \end{cases}

The exponent shift changes the expression for the first term, but not the ratio between consecutive terms. For example, in \(c r^{k-p}\), the term at \(k+1\) is \(r\) times the term at \(k\), whenever the terms are defined.

Why the Reindexing Identity Works

The substitution \(j=k-m\) gives a one-to-one correspondence between the original indices \(m,m+1,\ldots,n\) and the new indices \(0,1,\ldots,n-m\). At the corresponding indices, the terms agree: since \(k=m+j\),

c r^{k-p} = c r^{m+j-p} = \bigl(c r^{m-p}\bigr)r^j.

Thus the first term in the new expression is \(c r^{m-p}\), and each step in \(j\) multiplies the term by \(r\). Because the correspondence includes every original index exactly once, it neither omits nor duplicates terms. This proves the reindexing identity.

The finite formula follows from the same first-term representation. Let \(S=A+Ar+\cdots+Ar^{N-1}\). Multiplying by \(r\) shifts each term one place, so subtracting cancels all intermediate terms:

rS=Ar+Ar^2+\cdots+Ar^N, \qquad S-rS=A-Ar^N.

Therefore \((1-r)S=A(1-r^N)\). If \(r\ne1\), division by \(1-r\) gives \(S=A(1-r^N)/(1-r)\). When \(r=1\), division would be invalid; directly, \(S\) is the sum of \(N\) copies of \(A\), hence \(S=NA\).

Worked Examples

Worked Example: Shift a Finite Sum Beginning at a Nonzero Index

Rewrite and evaluate \(\displaystyle \sum_{k=3}^{7}2\left(\frac13\right)^{k-1}\).

First find the term at the starting index \(k=3\):

2\left(\frac13\right)^{3-1} = 2\left(\frac13\right)^2 = \frac{2}{9}.

The common ratio is \(1/3\). Set \(j=k-3\). The bounds become \(j=0\) and \(j=4\), so

\sum_{k=3}^{7}2\left(\frac13\right)^{k-1} = \frac29\sum_{j=0}^{4}\left(\frac13\right)^j.

There are five terms. Apply the finite formula:

\begin{aligned} \frac29\sum_{j=0}^{4}\left(\frac13\right)^j &=\frac29\cdot\frac{1-(1/3)^5}{1-1/3}\\ &=\frac29\cdot\frac{1-1/243}{2/3}\\ &=\frac29\cdot\frac{242}{243}\cdot\frac32\\ &=\frac{242}{729}. \end{aligned}

Check by listing the original terms with a common denominator:

\frac29+\frac{2}{27}+\frac{2}{81}+\frac{2}{243}+\frac{2}{729} = \frac{162+54+18+6+2}{729} = \frac{242}{729}.

The direct addition confirms the reindexed result.

Worked Example: Read the First Term from a Shifted Exponent

Rewrite and evaluate \(\displaystyle \sum_{k=4}^{8}7\left(\frac25\right)^{k-2}\).

At the lower bound \(k=4\), the exponent is \(4-2=2\). Hence the first term is

7\left(\frac25\right)^2 = 7\cdot\frac4{25} = \frac{28}{25}.

The common ratio is \(2/5\). With \(j=k-4\), the new bounds are \(0\) and \(4\), giving

\sum_{k=4}^{8}7\left(\frac25\right)^{k-2} = \frac{28}{25}\sum_{j=0}^{4}\left(\frac25\right)^j.

Evaluate the five-term sum:

\begin{aligned} \frac{28}{25}\sum_{j=0}^{4}\left(\frac25\right)^j &=\frac{28}{25}\cdot \frac{1-(2/5)^5}{1-2/5}\\ &=\frac{28}{25}\cdot \frac{1-32/3125}{3/5}\\ &=\frac{28}{25}\cdot\frac{3093}{3125}\cdot\frac53\\ &=\frac{28}{25}\cdot\frac{1031}{625}\\ &=\frac{28868}{15625}. \end{aligned}

For a check, the original terms are \(28/25,\;56/125,\;112/625,\;224/3125,\;448/15625\). Written with denominator \(15625\), their numerators are \(17500,\;7000,\;2800,\;1120,\;448\). Their sum is \(28868/15625\), as found.

Shifting an Infinite Geometric Series

An infinite geometric series also has a first term determined by its starting index. Consider \(\sum_{k=m}^{\infty} c r^{k-p}\). Replacing \(j\) by \(k-m\) in each finite partial sum gives the corresponding initial terms of a series beginning at \(j=0\), with first term \(A=c r^{m-p}\). If \(|r|<1\), the geometric series converges, and its value is \(A/(1-r)\).

Theorem: If \(A\) is the first term and \(|r|<1\), then \(\displaystyle \sum_{j=0}^{\infty}Ar^j=\frac{A}{1-r}\). Thus, when the terms are defined, \(\displaystyle \sum_{k=m}^{\infty}c r^{k-p}=\frac{c r^{m-p}}{1-r}\) under the same condition.

The condition concerns the common ratio, not the starting index. Shifting the starting index changes \(A\), but it does not change \(r\) or the condition \(|r|<1\). To justify the infinite-sum formula, the partial sum through \(j=N\) is \(A(1-r^{N+1})/(1-r)\). When \(|r|<1\), \(r^{N+1}\) tends to zero, so these partial sums tend to \(A/(1-r)\).

Worked Example: Shift an Infinite Series Starting at a Positive Exponent

Evaluate \(\displaystyle \sum_{n=4}^{\infty}\left(\frac35\right)^n\).

The first term is \((3/5)^4=81/625\), and the common ratio is \(3/5\). Since \(|3/5|<1\), the series converges. Set \(j=n-4\):

\sum_{n=4}^{\infty}\left(\frac35\right)^n = \frac{81}{625}\sum_{j=0}^{\infty}\left(\frac35\right)^j = \frac{81}{625}\cdot\frac{1}{1-3/5} = \frac{81}{625}\cdot\frac52 = \frac{81}{250}.

Check using the finite partial sum through \(n=4+N\). The shifted expression gives

\frac{81}{625}\cdot \frac{1-(3/5)^{N+1}}{1-3/5}.

As \(N\) tends to infinity, \((3/5)^{N+1}\) tends to zero, so this tends to \(\frac{81}{625}\cdot\frac52=\frac{81}{250}\), confirming the value.

A Common Indexing Mistake

A frequent error is to replace a sum beginning at \(k=m\) with one beginning at zero but leave the old coefficient unchanged. That can change the first term and therefore the entire sum.

Worked Example: Correct a Wrong Shift

Consider \(\displaystyle \sum_{k=3}^{6}4(1/2)^k\). A wrong calculation might shift the bounds to \(j=0,\ldots,3\) but keep \(4\) as the first term:

4\sum_{j=0}^{3}\left(\frac12\right)^j = 4\cdot\frac{1-(1/2)^4}{1-1/2} = 4\cdot\frac{15}{8} = \frac{15}{2}.

This cannot be correct: the original first term is \(4(1/2)^3=1/2\), not \(4\). The correct shift is

\begin{aligned} \sum_{k=3}^{6}4\left(\frac12\right)^k &=\frac12\sum_{j=0}^{3}\left(\frac12\right)^j\\ &=\frac12\cdot\frac{1-(1/2)^4}{1-1/2}\\ &=\frac12\cdot\frac{15}{8} =\frac{15}{16}. \end{aligned}

Checking the original terms gives \(1/2+1/4+1/8+1/16=15/16\). The correction is to recalculate the first term at the original lower bound before shifting.

A Reliable Checklist

When rewriting a geometric sum with a shifted index or exponent, use the following sequence:

Find the original lower bound.
Use it to identify which term is first.

Evaluate that term.
Substitute the lower index into the entire term, including its exponent.

Keep the common ratio.
Increasing the index by one multiplies each term by the same ratio.

Shift the bounds consistently.
For a finite sum from \(m\) to \(n\), the shifted bounds are \(0\) to \(n-m\), giving \(n-m+1\) terms.

Check the result.
Verify the first few terms or compare with the original partial sum.

These steps are useful because index notation can hide the actual terms. Once the first term, ratio, and number of terms are clear, the summation formula can be applied without guessing. This is especially important when a series begins at a later exponent: the omitted early powers are not part of the sum, so the first term must reflect that omission.

Check Your Understanding

For each question, identify the first term before applying a geometric-sum formula.

Rewrite \(\displaystyle \sum_{k=2}^{5}3(1/4)^{k-1}\) as a sum beginning at index zero. How many terms are there?
What is the first term and common ratio in \(\displaystyle \sum_{n=3}^{7}5(2/3)^n\)?
Under what condition on \(r\) does \(\displaystyle \sum_{k=m}^{\infty}c r^{k-p}\) converge by the geometric-series formula?
Explain why shifting a finite sum from indices \(m,\ldots,n\) to indices \(0,\ldots,n-m\) preserves the number of terms.
What error occurs if the bounds are shifted but the first term is not recalculated?