The Central Idea
Infinite series become manageable when their terms have structure. Two of the most important exact structures are
\[ a+ar+ar^2+ar^3+\cdots \]and
\[ (b_1-b_2)+(b_2-b_3)+(b_3-b_4)+\cdots. \]The first is geometric. The second is telescoping. Geometric structure is controlled by repeated multiplication. Telescoping structure is controlled by cancellation.
Why Partial Sums Still Come First
An infinite series is always defined through finite partial sums. For
\[ \sum_{n=1}^{\infty}a_n, \]write
\[ S_N=\sum_{n=1}^{N}a_n. \]If we can find a useful formula for \(S_N\), then the infinite sum is found by taking
\[ \lim_{N\to\infty}S_N. \]What Is a Geometric Sequence?
A sequence is geometric when each term is obtained from the previous term by multiplying by a fixed number \(r\), called the common ratio.
If the first term is \(a\), the sequence has the form
\[ a,\ ar,\ ar^2,\ ar^3,\ldots \]and its general term is
\[ \boxed{a_n=ar^{n-1}}. \]From a Geometric Sequence to a Geometric Series
Adding the terms of a geometric sequence produces a geometric series:
\[ \boxed{a+ar+ar^2+ar^3+\cdots}. \]In sigma notation,
\[ \boxed{\sum_{n=0}^{\infty}ar^n}. \]The starting index changes the exponent pattern, but the underlying idea is always repeated multiplication by the same ratio.
Finding the Common Ratio
When consecutive nonzero terms are known, divide a term by the preceding term:
\[ \boxed{r=\frac{a_{n+1}}{a_n}}. \]For
\[ 12,\ 6,\ 3,\ \frac32,\ldots, \]the common ratio is
\[ r=\frac6{12}=\frac12. \]Recognizing a Geometric Series
Consider
\[ 8-4+2-1+\frac12-\cdots. \]Every term is the previous term multiplied by
\[ -\frac12. \]Therefore the series is geometric with
\[ a=8, \qquad r=-\frac12. \]The Finite Geometric Sum
Before studying infinitely many terms, consider the finite sum
\[ S_N=a+ar+ar^2+\cdots+ar^{N-1}. \]When \(r\ne1\), this sum has the closed form
\[ \boxed{S_N=a\frac{1-r^N}{1-r}}. \]This formula is the key to the infinite geometric series.
Deriving the Finite Geometric Formula
Start with
\[ S_N=a+ar+ar^2+\cdots+ar^{N-1}. \]Multiply by \(r\):
\[ rS_N=ar+ar^2+\cdots+ar^{N-1}+ar^N. \]Subtract the second equation from the first:
\[ S_N-rS_N=a-ar^N. \]Thus
\[ S_N(1-r)=a(1-r^N), \]so, for \(r\ne1\),
\[ \boxed{S_N=a\frac{1-r^N}{1-r}}. \]The Infinite Geometric Sum
If \(|r|<1\), then
\[ r^N\to0. \]Therefore
\[ \lim_{N\to\infty}S_N =\lim_{N\to\infty}a\frac{1-r^N}{1-r} =\frac{a}{1-r}. \]Hence
\[ \boxed{\sum_{n=0}^{\infty}ar^n=\frac{a}{1-r}\quad\text{when }|r|<1.} \]The Geometric Convergence Rule
A nonzero geometric series
\[ a+ar+ar^2+\cdots \]converges exactly when
\[ \boxed{|r|<1}. \]If \(|r|\ge1\), the terms do not approach zero, except in the trivial case \(a=0\). Therefore the series diverges.
Why the Absolute Value of the Ratio Matters
The condition \(|r|<1\) means the magnitude of each successive term shrinks by a fixed factor.
Both
\[ r=\frac12 \]and
\[ r=-\frac12 \]satisfy the condition. A negative ratio changes signs, but the magnitudes still decay geometrically.
Case: \(0<r<1\)
When the ratio is positive and smaller than \(1\), all terms have the same sign and decrease in magnitude.
For example,
\[ 1+\frac12+\frac14+\frac18+\cdots \]has \(a=1\) and \(r=\frac12\), so
\[ \boxed{\sum_{n=0}^{\infty}\left(\frac12\right)^n=2}. \]Case: \(-1<r<0\)
A negative ratio produces alternating signs while magnitudes shrink.
For example,
\[ 1-\frac12+\frac14-\frac18+\cdots \]has \(a=1\) and \(r=-\frac12\), so
\[ \boxed{\frac{1}{1-(-1/2)}=\frac23}. \]Case: \(r=1\)
If \(r=1\), then every term equals \(a\):
\[ a+a+a+\cdots. \]Unless \(a=0\), the terms do not approach zero. Therefore the series diverges.
Case: \(r=-1\)
If \(r=-1\), then the terms alternate between \(a\) and \(-a\):
\[ a-a+a-a+\cdots. \]The partial sums oscillate and do not approach a single limit when \(a\ne0\). The series diverges.
Case: \(|r|>1\)
If the magnitude of the ratio exceeds \(1\), then the terms grow rather than shrink.
For example,
\[ 1+2+4+8+\cdots \]has \(r=2\). Since the terms do not approach zero, the series diverges.
Example: Direct Geometric Evaluation
Evaluate
\[ 5+\frac52+\frac54+\frac58+\cdots. \]Here
\[ a=5, \qquad r=\frac12. \]Since \(|r|<1\),
\[ \boxed{S=\frac5{1-1/2}=10}. \]Example: Alternating Geometric Evaluation
Evaluate
\[ 6-2+\frac23-\frac29+\cdots. \]The ratio is
\[ r=-\frac13. \]Thus
\[ \boxed{S=\frac6{1+1/3}=\frac92}. \]Example: A Geometric Series That Diverges
Consider
\[ 3-6+12-24+\cdots. \]The common ratio is \(-2\). Since
\[ |-2|=2\ge1, \]the series diverges.
The First Term Depends on the Starting Index
The formula
\[ \frac{a}{1-r} \]uses the actual first term of the series being summed.
For example,
\[ \sum_{n=3}^{\infty}\left(\frac12\right)^n \]begins with \(a=(1/2)^3=1/8\), not \(1\). Therefore
\[ \boxed{S=\frac{1/8}{1-1/2}=\frac14}. \]Geometric Series Written with \(n=1\)
A common form is
\[ \sum_{n=1}^{\infty}ar^{n-1}. \]Substituting \(n=1\) gives the first term \(a\), so
\[ \boxed{\sum_{n=1}^{\infty}ar^{n-1}=\frac{a}{1-r},\quad |r|<1.} \]Geometric Series Written with \(n=0\)
Another common form is
\[ \sum_{n=0}^{\infty}ar^n. \]Since \(r^0=1\), the first term is again \(a\). Thus the same formula applies:
\[ \boxed{\sum_{n=0}^{\infty}ar^n=\frac{a}{1-r},\quad |r|<1.} \]Factoring to Reveal a Geometric Series
A geometric pattern may be easier to see after extracting a constant. Consider
\[ \sum_{n=0}^{\infty}\frac{3}{5^{n+1}}. \]Rewrite it as
\[ \frac35\sum_{n=0}^{\infty}\left(\frac15\right)^n. \]Therefore
\[ \boxed{S=\frac35\cdot\frac1{1-1/5}=\frac34}. \]Using Algebra Before the Formula
Consider
\[ \sum_{n=1}^{\infty}\frac{2^{n+1}}{5^n}. \]Rewrite the term:
\[ \frac{2^{n+1}}{5^n}=2\left(\frac25\right)^n. \]The first term occurs at \(n=1\), so it is \(4/5\), and the common ratio is \(2/5\). Hence
\[ \boxed{S=\frac{4/5}{1-2/5}=\frac43}. \]Geometric Series and Repeating Decimals
Repeating decimals can be written as geometric series. For example,
\[ 0.777\ldots =\frac7{10}+\frac7{100}+\frac7{1000}+\cdots. \]Here
\[ a=\frac7{10}, \qquad r=\frac1{10}. \]Therefore
\[ \boxed{0.777\ldots=\frac{7/10}{1-1/10}=\frac79}. \]Repeating Blocks of Digits
A repeating block can also generate a geometric series. For example,
\[ 0.121212\ldots =\frac{12}{100}+\frac{12}{100^2}+\frac{12}{100^3}+\cdots. \]Thus
\[ \boxed{0.121212\ldots=\frac{12/100}{1-1/100}=\frac{12}{99}=\frac4{33}}. \]The Remainder of a Convergent Geometric Series
If a geometric series converges to \(S\), the error after the first \(N\) terms is
\[ R_N=S-S_N. \]For
\[ a+ar+ar^2+\cdots, \]the remaining tail begins with \(ar^N\), so
\[ \boxed{R_N=\frac{ar^N}{1-r}}, \qquad |r|<1. \]Geometric Remainder as an Error Estimate
The magnitude of the geometric remainder is
\[ \boxed{|R_N|=\frac{|a|\,|r|^N}{|1-r|}}. \]Because \(|r|^N\to0\), this error can be made as small as desired by using enough terms.
Example: How Much Remains?
For
\[ 1+\frac12+\frac14+\frac18+\cdots, \]the sum is \(2\). After four terms,
\[ S_4=1+\frac12+\frac14+\frac18=\frac{15}{8}. \]Therefore
\[ \boxed{R_4=2-\frac{15}{8}=\frac18}. \]The same result comes from \(R_4=r^4/(1-r)\).
Solving for an Unknown Ratio
Suppose a convergent geometric series has first term \(6\) and sum \(10\). Then
\[ 10=\frac6{1-r}. \]Solving,
\[ 10-10r=6, \] \[ r=\frac25. \]Since \(|2/5|<1\), the result is consistent with convergence.
Geometric Models of Repeated Retention
Geometric series appear whenever a fixed fraction of something remains or repeats at each stage. If an initial contribution is \(A\) and each later contribution is a fraction \(r\) of the previous one, the total accumulated contribution is
\[ A+Ar+Ar^2+\cdots. \]When \(|r|<1\), the total is
\[ \boxed{\frac{A}{1-r}}. \]Geometric Series Are Exact, Not Merely Approximate
A statement such as
\[ 1+\frac12+\frac14+\cdots=2 \]does not mean that some finite partial sum literally equals \(2\). It means the partial sums approach \(2\) as their number of terms grows without bound.
What Is a Telescoping Series?
A telescoping series is a series whose partial sums simplify because many terms cancel.
A standard pattern is
\[ \sum_{n=1}^{\infty}(b_n-b_{n+1}). \]Writing the first few terms exposes the cancellation.
The Basic Telescoping Pattern
Expand
\[ (b_1-b_2)+(b_2-b_3)+(b_3-b_4)+\cdots. \]The \(-b_2\) cancels with \(+b_2\), the \(-b_3\) cancels with \(+b_3\), and so on. The finite partial sum is
\[ \boxed{S_N=b_1-b_{N+1}}. \]Why We Telescope the Partial Sum
Cancellation is a finite algebraic operation. We first write
\[ S_N=\sum_{n=1}^{N}(b_n-b_{n+1}), \]perform the cancellation, and only then take \(N\to\infty\). This keeps the logic grounded in ordinary finite sums.
The Standard Telescoping Sum Formula
If
\[ S_N=b_1-b_{N+1} \]and \(b_{N+1}\to L\), then
\[ \boxed{\sum_{n=1}^{\infty}(b_n-b_{n+1})=b_1-L}. \]In many common examples, \(L=0\), leaving simply \(b_1\).
Example: The Classic Telescoping Series
Consider
\[ \sum_{n=1}^{\infty}\frac{1}{n(n+1)}. \]Use partial fractions:
\[ \frac1{n(n+1)}=\frac1n-\frac1{n+1}. \]Therefore
\[ S_N=1-\frac1{N+1}, \]and hence
\[ \boxed{\sum_{n=1}^{\infty}\frac1{n(n+1)}=1}. \]Writing Out Terms Reveals the Cancellation
The previous series becomes
\[ \left(1-\frac12\right) +\left(\frac12-\frac13\right) +\left(\frac13-\frac14\right) +\cdots. \]Everything in the interior cancels, leaving the first positive term and the final negative term of the finite partial sum.
Partial Fractions as a Telescoping Tool
Rational expressions often telescope only after decomposition. A common pattern is
\[ \frac{1}{n(n+k)}. \]Since
\[ \frac1{n(n+k)} =\frac1k\left(\frac1n-\frac1{n+k}\right), \]the series may telescope with a shift of \(k\) positions.
Example: A Two-Step Telescope
Evaluate
\[ \sum_{n=1}^{\infty}\frac1{n(n+2)}. \]Decompose:
\[ \frac1{n(n+2)}=\frac12\left(\frac1n-\frac1{n+2}\right). \]Then
\[ S_N=\frac12\left(1+\frac12-\frac1{N+1}-\frac1{N+2}\right). \]Taking the limit gives
\[ \boxed{\frac34}. \]Not Every Telescoping Series Leaves Only Two Terms
When the cancellation skips more than one position, several initial and final terms may remain. The important point is not that exactly two terms survive, but that the number of uncanceled terms stays manageable as \(N\) grows.
Example: Shifted Telescoping
Consider
\[ \sum_{n=3}^{\infty}\left(\frac1n-\frac1{n+1}\right). \]The partial sum is
\[ S_N=\frac13-\frac1{N+1}. \]Therefore
\[ \boxed{\sum_{n=3}^{\infty}\left(\frac1n-\frac1{n+1}\right)=\frac13}. \]Telescoping with Square Roots
Some expressions telescope after rationalization. Consider
\[ \frac1{\sqrt{n+1}+\sqrt n}. \]Multiply numerator and denominator by the conjugate:
\[ \frac1{\sqrt{n+1}+\sqrt n} =\sqrt{n+1}-\sqrt n. \]This reveals a telescoping difference.
A Telescoping Series Can Still Diverge
Telescoping describes cancellation; it does not automatically guarantee a finite limit. For example,
\[ \sum_{n=1}^{\infty}(\sqrt{n+1}-\sqrt n) \]has partial sums
\[ S_N=\sqrt{N+1}-1, \]which grow without bound. Therefore the series diverges.
The Surviving Boundary Terms Determine the Result
After cancellation, a telescoping series is controlled by the terms that remain at the beginning and end of the partial sum. Convergence therefore depends on what happens to the trailing boundary terms as \(N\to\infty\).
Example: A Nonzero Limiting Boundary
Suppose
\[ b_n=2+\frac1n. \]Then
\[ \sum_{n=1}^{\infty}(b_n-b_{n+1}) \]has partial sums
\[ S_N=b_1-b_{N+1}=3-\left(2+\frac1{N+1}\right). \]Thus
\[ \boxed{S=1}. \]Recognizing a Hidden Difference
A telescoping series may not initially be written as \(b_n-b_{n+1}\). The central task is often to rewrite the term so that successive copies of the same expression appear with opposite signs.
Useful tools include partial fractions, conjugates, algebraic splitting, and index shifts.
Geometric and Telescoping Structure Can Overlap
Consider
\[ \sum_{n=0}^{\infty}\left(\frac1{2^n}-\frac1{2^{n+1}}\right). \]The terms telescope, but each difference also equals
\[ \frac1{2^{n+1}}, \]which forms a geometric series. Different valid structures can lead to the same result.
Two Ways to Evaluate the Same Series
For
\[ \sum_{n=0}^{\infty}\left(\frac1{2^n}-\frac1{2^{n+1}}\right), \]telescoping gives
\[ S_N=1-\frac1{2^{N+1}}\to1. \]Geometrically, the term is \(2^{-(n+1)}\), so
\[ S=\frac{1/2}{1-1/2}=1. \]Agreement between methods provides a useful check.
Geometric Versus Telescoping: What to Look For
- Geometric: consecutive terms differ by a constant multiplicative ratio.
- Telescoping: expanded partial sums contain repeated terms with opposite signs.
- Geometric methods focus on \(a\) and \(r\).
- Telescoping methods focus on rewriting and cancellation.
- Both methods ultimately rely on a limit of finite partial sums.
Finite Changes Do Not Destroy the Basic Method
A series may begin with a few exceptional terms and then become geometric or telescoping. Separate the finite beginning from the structured tail. The finite portion can be added directly, while the infinite tail is handled by its appropriate method.
Example: A Finite Beginning Plus a Geometric Tail
Suppose
\[ 7+3+\frac32+\frac34+\frac38+\cdots. \]The tail beginning at \(3\) is geometric with ratio \(1/2\), so
\[ 3+\frac32+\frac34+\cdots=\frac3{1-1/2}=6. \]Therefore the full sum is
\[ \boxed{7+6=13}. \]Example: A Finite Beginning Plus a Telescoping Tail
Consider
\[ 4+\sum_{n=2}^{\infty}\left(\frac1n-\frac1{n+1}\right). \]The telescoping tail sums to \(1/2\). Hence the full expression equals
\[ \boxed{\frac92}. \]Common Error: Using the Geometric Formula Without Checking \(|r|<1\)
The algebraic expression
\[ \frac{a}{1-r} \]is the infinite sum only when the geometric series converges. For
\[ 1+2+4+8+\cdots, \]inserting \(r=2\) into the formula would produce \(-1\), but the actual series diverges. The convergence condition is not optional.
Common Error: Using the Wrong First Term
In
\[ \sum_{n=2}^{\infty}\left(\frac13\right)^n, \]the first term is
\[ \left(\frac13\right)^2=\frac19, \]not \(1\) and not \(1/3\). Always evaluate the term at the actual starting index.
Common Error: Confusing the Ratio with the First Term
In
\[ 12+4+\frac43+\cdots, \]the first term is \(12\), while the ratio is \(1/3\). These play different roles in
\[ S=\frac{a}{1-r}. \]Common Error: Assuming Terms Approaching Zero Proves a Geometric Sum
A series is geometric only when consecutive terms have a constant ratio. The fact that its terms approach zero does not make it geometric and does not by itself prove convergence.
Common Error: Cancelling Across an Infinite Expression Informally
Telescoping should be performed on the finite partial sum \(S_N\). Write a finite number of terms, cancel them algebraically, obtain a formula in \(N\), and then take the limit. This avoids treating infinity as though it were a final index.
Common Error: Missing Surviving Terms
In a shifted telescope such as
\[ \sum_{n=1}^{N}\left(\frac1n-\frac1{n+2}\right), \]two positive beginning terms and two negative ending terms survive. Do not assume every telescope reduces to exactly one first term and one last term.
Common Error: Declaring Every Telescoping Series Convergent
Cancellation may simplify the partial sum without producing a finite limit. After telescoping, the final step is still to evaluate
\[ \lim_{N\to\infty}S_N. \]A Reliable Procedure for Geometric Series
- Write several terms if the pattern is not immediately clear.
- Identify the actual first term \(a\).
- Find the common ratio \(r\).
- Verify that the ratio is constant.
- Check whether \(|r|<1\).
- If \(|r|<1\), use \(S=a/(1-r)\).
- If \(|r|\ge1\), conclude divergence unless the series is trivial.
- Check the sign and approximate size of the result for reasonableness.
A Reliable Procedure for Telescoping Series
- Look for a difference of neighboring or shifted expressions.
- Rewrite the term when needed using partial fractions, conjugates, or algebra.
- Form the finite partial sum \(S_N\).
- Write enough terms to expose the cancellation pattern.
- Cancel only terms that genuinely appear with opposite signs.
- Identify all surviving boundary terms.
- Simplify \(S_N\) as a function of \(N\).
- Take \(N\to\infty\).
- State convergence and the sum only if the limit is finite.
Choosing Between the Two Methods
Before doing heavy algebra, ask two quick questions:
- Do consecutive terms have a constant ratio?
- Can the general term be rewritten as a difference that will cancel?
A series may answer yes to one, both, or neither. Pattern recognition is part of the mathematics, not merely a shortcut.
Practice 1: Identify a Geometric Series
Determine whether the series is geometric. If so, identify \(a\) and \(r\):
\[ 9-3+1-\frac13+\cdots. \]Practice 2: Evaluate a Geometric Series
Evaluate
\[ 4+1+\frac14+\frac1{16}+\cdots. \]Practice 3: Test Geometric Convergence
Determine whether
\[ \sum_{n=0}^{\infty}5\left(-\frac43\right)^n \]converges or diverges.
Practice 4: Starting Index
Evaluate
\[ \sum_{n=4}^{\infty}\left(\frac12\right)^n. \]Practice 5: Geometric Remainder
For
\[ 3+1+\frac13+\frac19+\cdots, \]find the exact remainder after the first four terms.
Practice 6: Basic Telescoping
Evaluate
\[ \sum_{n=1}^{\infty}\left(\frac1n-\frac1{n+1}\right). \]Practice 7: Partial Fractions and Telescoping
Evaluate
\[ \sum_{n=1}^{\infty}\frac1{(n+1)(n+2)}. \]Practice 8: Two-Step Telescoping
Evaluate
\[ \sum_{n=1}^{\infty}\frac1{n(n+3)}. \]Practice 9: Telescoping Divergence
Determine whether
\[ \sum_{n=1}^{\infty}(\sqrt{n+2}-\sqrt{n+1}) \]converges or diverges.
Practice 10: Identify the Error
A student evaluates
\[ 2+6+18+54+\cdots \]by writing
\[ \frac{2}{1-3}=-1. \]Explain why this conclusion is invalid.
Practice Answers
Practice 1
Yes. The ratio is constant:
\[ r=-\frac13. \]Thus
\[ \boxed{a=9,\qquad r=-\frac13}. \]Practice 2
Here \(a=4\) and \(r=1/4\). Therefore
\[ \boxed{S=\frac4{1-1/4}=\frac{16}{3}}. \]Practice 3
The ratio is \(-4/3\), so
\[ \left|-\frac43\right|>1. \]Therefore
\[ \boxed{\text{the series diverges}.} \]Practice 4
The first term is \((1/2)^4=1/16\), and \(r=1/2\). Hence
\[ \boxed{S=\frac{1/16}{1-1/2}=\frac18}. \]Practice 5
Here \(a=3\), \(r=1/3\), and \(N=4\). Therefore
\[ R_4=\frac{3(1/3)^4}{1-1/3} =\boxed{\frac1{18}}. \]Practice 6
\[ S_N=1-\frac1{N+1}. \]Thus
\[ \boxed{S=1}. \]Practice 7
Since
\[ \frac1{(n+1)(n+2)}=\frac1{n+1}-\frac1{n+2}, \]the series telescopes to
\[ \boxed{\frac12}. \]Practice 8
Decompose:
\[ \frac1{n(n+3)}=\frac13\left(\frac1n-\frac1{n+3}\right). \]Three initial terms survive in the limit:
\[ S=\frac13\left(1+\frac12+\frac13\right) =\boxed{\frac{11}{18}}. \]Practice 9
The partial sum is
\[ S_N=\sqrt{N+2}-\sqrt2. \]Since \(S_N\to\infty\),
\[ \boxed{\text{the series diverges}.} \]Practice 10
The series is geometric with \(r=3\), but the infinite geometric sum formula applies only when \(|r|<1\). Since \(|3|>1\), the terms grow and the series diverges. The value \(-1\) is not its sum.
What to Remember
A geometric series has the form
\[ a+ar+ar^2+\cdots. \]It converges exactly when
\[ \boxed{|r|<1}, \]and then
\[ \boxed{S=\frac{a}{1-r}}. \]A telescoping series is recognized by cancellation in its partial sums. The standard difference pattern is
\[ \sum_{n=1}^{\infty}(b_n-b_{n+1}), \]whose partial sums satisfy
\[ \boxed{S_N=b_1-b_{N+1}}. \]- Always identify the actual first term of a geometric series.
- Check \(|r|<1\) before using the infinite geometric sum formula.
- Use finite partial sums to justify telescoping cancellation.
- Partial fractions and conjugates can reveal hidden telescoping structure.
- Not every telescoping series converges; the surviving boundary terms must have a finite limit.
- When more than one structure is available, use the clearest one and cross-check when useful.
Where We Go Next
Geometric and telescoping series are unusually cooperative because their partial sums can often be simplified exactly. Most positive-term series do not present such an obvious closed form.
The next lesson, Positive-Term Convergence Tests, develops a systematic toolbox for deciding whether positive series converge even when their sums cannot be evaluated directly. Comparison, limit comparison, integral, ratio, and root ideas will extend the structural thinking used here into a broader convergence framework.