Choosing a Sum Strategy
A sum written with sigma notation can look more complicated than the calculation it represents. The useful first step is to identify its structure. If each term is obtained by multiplying the preceding term by a fixed ratio, the sum is geometric. If terms are written as differences and repeated pieces cancel, the sum is telescoping.
In either case, the endpoints matter. A finite sum from \(k=m\) to \(k=n\) contains exactly \(n-m+1\) terms. The first index determines the first term, and the last index determines the final term. Writing those terms explicitly before applying a formula is a reliable way to avoid an off-by-one error.
The finite geometric formula applies to a fixed number of terms. For an infinite geometric series, the sum is defined through the limit of its partial sums, and a finite limit is obtained when the common ratio satisfies \(|r|<1\). Telescoping sums are handled by displaying enough terms to see which terms cancel and which remain.
A Useful Geometric-Sum Identity
For \(N\) terms with first term \(a\) and ratio \(r\), the finite geometric-sum formula is:
Here \(N\) is a positive integer. If the sum is written with index \(k=m\) at the start and \(k=n\) at the end, the number of terms is \(N=n-m+1\). The first term is the summand evaluated at \(k=m\), not necessarily the coefficient written outside the sum.
The identity follows by subtracting two expressions for the same sum. Let \(S=a+ar+\cdots+ar^{N-1}\). Multiplication by \(r\) shifts the terms one place, so \(rS=ar+ar^2+\cdots+ar^N\). Subtracting cancels the shared middle terms. When \(r\ne1\), division by \(1-r\) gives the formula:
If \(r=1\), every one of the \(N\) terms equals \(a\), so the sum is \(Na\). The formula with a denominator \(1-r\) cannot be used in that case.
Worked Examples
Worked Example: A Finite Geometric Sum With a Shifted Index
Evaluate \(\displaystyle\sum_{k=2}^{5}5\cdot2^{k-1}\). First count the terms: \(N=5-2+1=4\). The first term is found by substituting the lower endpoint \(k=2\): \(5\cdot2^{2-1}=10\). Each step increases the exponent by one, so the common ratio is \(2\).
The four terms are \(10,20,40,80\). Apply the finite formula with \(a=10\), \(r=2\), and \(N=4\):
Check by adding the listed terms: \(10+20=30\), \(30+40=70\), and \(70+80=150\). The formula and direct sum agree. In particular, the lower endpoint \(k=2\) gives \(10\), not \(5\); using \(5\) as the first term would shift the whole sequence.
Worked Example: An Infinite Geometric Series With Alternating Signs
Evaluate \(\displaystyle\sum_{n=0}^{\infty}6\left(-\frac14\right)^n\). The first term is \(6\), the common ratio is \(-\frac14\), and \(\left|-\frac14\right|=\frac14<1\). Thus the partial sums have a finite limit, and the infinite geometric sum is \(a/(1-r)\):
To check the result from partial sums, the first four terms are \(6,-\frac32,\frac38,-\frac{3}{32}\). Their sum is \(6-\frac32=\frac92\), then \(\frac92+\frac38=\frac{39}{8}\), then \(\frac{39}{8}-\frac{3}{32}=\frac{153}{32}\). For \(N\) terms, the finite partial sum is \(S_N=6\bigl(1-(-1/4)^N\bigr)/(1+1/4)\). Since \((-1/4)^N\) tends to zero, \(S_N\) tends to \(6/(5/4)=24/5\), as claimed.
The partial sum is not itself exactly \(24/5\) after four terms: \(\frac{153}{32}\ne\frac{24}{5}\). It approaches that value as the number of terms grows. The condition \(|r|<1\) is essential to this infinite-sum calculation.
Worked Example: A Telescoping Sum With Two Residual Terms at Each End
Evaluate \(\displaystyle\sum_{k=1}^{5}\left(\frac1k-\frac1{k+2}\right)\). Expand the five summands, keeping both fractions in each one:
The negative \(-\frac13,-\frac14,-\frac15\) cancel the positive \(\frac13,\frac14,\frac15\). What remains is:
Check by adding the original five summands separately: they are \(\frac23,\frac14,\frac{2}{15},\frac1{12},\frac{2}{35}\). Using denominator \(420\), their sum is \(\frac{280+105+56+35+24}{420}=\frac{500}{420}=\frac{25}{21}\). This agrees with the cancellation calculation and confirms the endpoints.
Worked Example: A Shifted Telescoping Sum With Three Terms Left at Each End
Evaluate \(\displaystyle\sum_{k=2}^{6}\left(\frac1{k-1}-\frac1{k+2}\right)\). The lower index is \(k=2\), so the first summand is \(1-\frac14\). The upper index is \(k=6\), so the last summand is \(\frac15-\frac18\). Expanding all five terms gives:
The \(\frac14\) and \(\frac15\) terms cancel across the sum. The remaining terms are the first three positive fractions and the last three negative fractions:
For a direct check, the five original summands are \(\frac34,\frac{3}{10},\frac16,\frac{3}{28},\frac{3}{40}\). With denominator \(840\), their sum is \(\frac{630+252+140+90+63}{840}=\frac{1175}{840}=\frac{235}{168}\). Both the direct calculation and the cancellation pattern give the same value.
How to Keep the Endpoints Straight
For a geometric sum, write down the first term, ratio, and number of terms before substituting into the formula. For a telescoping sum, substitute the first and last index into the summand and expand enough terms to display the cancellation. These steps make the limits of the sum visible instead of relying on a guess about what “the beginning” or “the end” should be.
For integer indices from \(m\) through \(n\), inclusive, use \(n-m+1\).
Check whether consecutive terms have a constant ratio, or whether each summand is a difference with overlapping terms.
Apply the finite geometric identity, the infinite geometric limit when \(|r|<1\), or cancellation of matching terms.
For a finite sum, expand or add the terms. For an infinite geometric sum, verify the partial-sum limit and its ratio condition.
A Common Endpoint Mistake
Consider \(\displaystyle\sum_{k=1}^{4}\left(\frac1k-\frac1{k+1}\right)\). A common mistake is to write the surviving negative endpoint as \(-\frac14\), forgetting that the negative fraction in the final summand has denominator \(4+1=5\). The incorrect calculation \(1-\frac14=\frac34\) omits the actual last term.
Expanding the four summands shows the correct endpoint:
The check is direct: the four summands are \(\frac12,\frac16,\frac1{12},\frac1{20}\). With denominator \(60\), their sum is \(\frac{30+10+5+3}{60}=\frac{48}{60}=\frac45\). Substituting the upper index into the original summand is the safeguard: at \(k=4\), the negative term is \(-1/(4+1)\), not \(-1/4\).
Why These Checks Matter
Geometric and telescoping sums can both produce a short final expression, but the short expression is trustworthy only when it represents exactly the terms specified by the indices. For a geometric sum, an incorrect first term or term count changes the powers in the formula. For a telescoping sum, cancellation can hide an endpoint that was never included or remove a term that should remain.
In practice, a compact check is often enough: list the first and last terms, verify the number of terms, and compare the result with a direct expansion on a small finite sum. For an infinite geometric series, the check must also show that the partial sums approach the claimed value under the condition \(|r|<1\). This keeps the calculation tied to the definition of the sum rather than to a memorized pattern.
Check Your Understanding
For each question, pay particular attention to the first and last terms before calculating.
- In \(\displaystyle\sum_{k=3}^{7}2\cdot3^{k-3}\), what are the first term, common ratio, and number of terms?
- Evaluate \(\displaystyle\sum_{j=0}^{4}4\left(\frac12\right)^j\) using the finite geometric-sum formula, then check by listing the terms.
- Does \(\displaystyle\sum_{n=0}^{\infty}5\left(-\frac13\right)^n\) have a finite geometric sum? If so, find it and state the condition that permits the calculation.
- Expand \(\displaystyle\sum_{k=1}^{4}\left(\frac1k-\frac1{k+2}\right)\). Which terms remain after cancellation?
- For \(\displaystyle\sum_{k=2}^{5}\left(\frac1{k-1}-\frac1k\right)\), identify the first and last summands before simplifying.
