Geometric Series With Shifted Indices and Starting Terms

Why the Starting Index Matters

A geometric series is built from terms with a constant ratio between consecutive terms. When its index begins at a nonzero integer, the first term may not be the coefficient written in the formula. Keeping track of that difference is essential: the common ratio determines the pattern, but the actual first term determines the scale of the sum.

For example, in \(\sum_{k=3}^{\infty}4(1/2)^k\), the coefficient is \(4\), but the first term is \(4(1/2)^3=1/2\). Treating \(4\) as the first term would give the wrong sum. The reliable approach is to identify the first term explicitly, retain the common ratio, and then apply the geometric sum formula.

Two Ways to Write a Shifted Geometric Series

Definition: A geometric series has terms formed by multiplying each term by a constant common ratio \(r\). If the index begins at \(k=m\), a common form is \(\sum_{k=m}^{\infty}c r^k\). Here \(c\) is the coefficient in the formula; the first term is \(c r^m\).

There is another useful way to write the same pattern. If the first term is \(A\), and the index begins at \(k=m\), write the terms as \(A r^{k-m}\). The exponent is zero at the first index, so this form makes the first term visible immediately:

\sum_{k=m}^{\infty} A r^{k-m} = A+Ar+Ar^2+\cdots.

These forms encode the same structure but make different information explicit. In \(c r^k\), the coefficient \(c\) is given, and the starting index determines the first term \(c r^m\). In \(A r^{k-m}\), the first term \(A\) is given directly. If the expressions describe the same series, then \(A=c r^m\).

Changing the index label does not change the terms or their order. Set \(j=k-m\). As \(k\) runs through \(m,m+1,m+2,\ldots\), \(j\) runs through \(0,1,2,\ldots\), and \(c r^k=c r^m r^j\). Thus a shifted series can be rewritten as a series beginning at index zero, with first term \(c r^m\).

The Finite Sum and Its Limit

A series is defined through its partial sums: the \(N\)th partial sum is the sum of its first \(N\) terms. For a geometric series beginning at \(k=m\), those terms have indices \(m\) through \(m+N-1\). When \(r\ne 1\), the finite geometric-sum formula gives:

\sum_{k=m}^{m+N-1}c r^k = c r^m\frac{1-r^N}{1-r}.

The factor \(c r^m\) is the first term. The \(N\) powers after it are \(1,r,\ldots,r^{N-1}\). If \(|r|<1\), then \(r^N\to 0\) as \(N\to\infty\), so the partial sums approach a finite limit.

Theorem: If \(|r|<1\), then \(\sum_{k=m}^{\infty}c r^k\) converges and has sum \(\frac{c r^m}{1-r}\). Equivalently, a geometric series with first term \(A\) and common ratio \(r\) has sum \(\frac{A}{1-r}\), provided \(|r|<1\).

The condition on the ratio is necessary for this infinite-sum formula. If \(r=1\) and the first term is nonzero, every term equals that first term. If \(r=-1\), the terms alternate between the first term and its negative, and the partial sums do not settle to a single limit. If \(|r|>1\) and the first term is nonzero, the terms do not approach zero. In each of these cases, the series does not converge. A zero first term is an exception: every term is zero, so the series sums to zero.

Why the Shifted Formula Works

The finite formula shows both the effect of the shift and the role of the convergence condition. The following derivation starts from the partial sum and explains how the formula arises.

Factor out the first term.
Write the \(N\) terms starting at \(k=m\) as \(c r^m(1+r+\cdots+r^{N-1})\).

Use the finite geometric identity.
For \(r\ne 1\), multiplying \(1+r+\cdots+r^{N-1}\) by \(1-r\) cancels all the intermediate powers, leaving \(1-r^N\).

Take the limit.
When \(|r|<1\), \(r^N\to 0\), so the partial sums approach \(c r^m/(1-r)\).

For completeness, the cancellation in the second step is:

\begin{aligned} (1-r)(1+r+\cdots+r^{N-1}) &=(1+r+\cdots+r^{N-1})-(r+r^2+\cdots+r^N)\\ &=1-r^N. \end{aligned}

Dividing by \(1-r\) gives the finite identity when \(r\ne 1\). Substituting it into the partial sum yields \(c r^m(1-r^N)/(1-r)\). Finally, if \(|r|<1\), taking \(N\to\infty\) makes \(r^N\) tend to zero and proves the stated infinite-sum formula. This also shows why the index shift affects the sum through the first term \(c r^m\), not by changing the denominator \(1-r\).

Worked Examples

Worked Example: A Coefficient With a Shifted Index

Evaluate \(\sum_{k=3}^{\infty}4(1/2)^k\).

The common ratio is \(r=1/2\), and the first index is \(m=3\). The first term is

4\left(\frac12\right)^3 =4\cdot\frac18 =\frac12.

Since \(|1/2|<1\), the series converges. Using the first-term form of the sum formula gives

\sum_{k=3}^{\infty}4\left(\frac12\right)^k = \frac{\frac12}{1-\frac12} = \frac{\frac12}{\frac12} =1.

Check: The \(N\)th partial sum is \(\frac12\frac{1-(1/2)^N}{1-1/2}=1-(1/2)^N\). As \(N\to\infty\), this tends to \(1\), confirming the answer. The first terms, \(1/2,1/4,1/8,\ldots\), also have the stated first term and ratio.

Worked Example: A Given First Term and Negative Ratio

Evaluate \(\sum_{n=4}^{\infty}7(-1/3)^{n-4}\).

This expression directly gives the first term \(A=7\), because the exponent is zero when \(n=4\). The common ratio is \(r=-1/3\), and \(|r|<1\). Therefore,

\sum_{n=4}^{\infty}7\left(-\frac13\right)^{n-4} = \frac{7}{1-(-\frac13)} = \frac{7}{1+\frac13} = \frac{7}{\frac43} = \frac{21}{4}.

Check: The \(N\)th partial sum is \(7\frac{1-(-1/3)^N}{1+1/3}=\frac{21}{4}\bigl(1-(-1/3)^N\bigr)\). Since \((-1/3)^N\to0\), the partial sums tend to \(21/4\). The first terms are \(7,-7/3,7/9,\ldots\), confirming the first term and the negative ratio.

Worked Example: A Scaled Coefficient With a Shifted Exponent

Evaluate \(\sum_{j=2}^{\infty}-6(3/5)^j\).

Here the coefficient is \(-6\), but the first term is \(-6(3/5)^2\). Compute it before applying the sum formula:

-6\left(\frac35\right)^2 = -6\cdot\frac9{25} = -\frac{54}{25}.

The common ratio is \(3/5\), so \(|3/5|<1\). The sum is

\sum_{j=2}^{\infty}-6\left(\frac35\right)^j = \frac{-\frac{54}{25}}{1-\frac35} = \frac{-\frac{54}{25}}{\frac25} = -\frac{54}{25}\cdot\frac52 = -\frac{27}{5}.

Check: The \(N\)th partial sum equals \(-\frac{54}{25}\frac{1-(3/5)^N}{1-3/5}=-\frac{27}{5}\bigl(1-(3/5)^N\bigr)\). Its limit is \(-27/5\). The terms begin \(-54/25,-162/125,\ldots\); each term is \(3/5\) times the preceding one, as required.

A Common Indexing Mistake

A frequent error is to use the coefficient as the first term, ignoring the starting index. For instance, one might incorrectly calculate

\sum_{k=3}^{\infty}4\left(\frac12\right)^k \overset{\text{wrong}}{=} \frac{4}{1-\frac12} =8.

This applies the formula as though the series began with \(4\), but its first term is \(4(1/2)^3=1/2\). The correction is to include the factor contributed by the starting index:

\sum_{k=3}^{\infty}4\left(\frac12\right)^k = \frac{4(1/2)^3}{1-\frac12} = \frac{\frac12}{\frac12} =1.

The wrong result \(8\) is not a small arithmetic slip; it sums a different geometric series, one whose first term is \(4\). Always calculate the first term from the actual starting index, or rewrite the exponent so that it is zero at that index.

Using the Formula Reliably

For a series written with a shifted index, keep the following quantities separate:

Quantity	How to identify it
Starting index	The first value taken by the summation index
Coefficient	The constant factor multiplying the power of the ratio
First term	Substitute the starting index into the term formula
Common ratio	The factor taking one term to the next
Convergence condition	For a nonzero first term, check that \(\|r\|<1\)

Then use the first term divided by \(1-r\). This procedure applies whether the series is written as \(c r^k\), with a coefficient and an explicit starting index, or as \(A r^{k-m}\), with the first term shown directly. A shift changes which power appears first; scaling changes the size of that first term. Neither changes the common ratio.

This distinction is useful whenever a series formula is given in a form that does not begin at index zero or one. It prevents index bookkeeping from being mistaken for a change in the geometric pattern, and it makes clear why two series with the same ratio can have different sums: their first terms may differ.

Check Your Understanding

For each question, identify the first term and ratio before deciding which sum formula applies.

For \(\sum_{k=2}^{\infty}3(1/4)^k\), what are the first term and common ratio, and what is the sum?
Rewrite \(\sum_{n=5}^{\infty}c r^n\) in a form that displays its first term and begins the exponent at zero.
Evaluate \(\sum_{j=3}^{\infty}5(-1/2)^{j-3}\). What condition allows the infinite-sum formula?
For a nonzero first term and common ratio \(r=1\), do the partial sums converge? Explain briefly.
A student uses \(c/(1-r)\) for \(\sum_{k=m}^{\infty}c r^k\). What factor has the student omitted?