Three Objects to Classify
A sequence of terms, its associated infinite series, and the sequence of partial sums are closely related, but they are not interchangeable. A reliable classification begins by identifying which object is being discussed.
Let \(a_1,a_2,a_3,\ldots\) be a sequence. Its associated series is written \(\sum_{n=1}^{\infty} a_n\). For each positive integer \(N\), its \(N\)th partial sum is \(S_N=a_1+a_2+\cdots+a_N\). Thus the partial sums themselves form a sequence \(S_1,S_2,S_3,\ldots\).
This definition makes the relationship precise: the series and its partial-sum sequence have the same convergence behavior by definition. The sequence of terms is a separate object. Its behavior must be classified on its own, even though its terms are what we add to make the partial sums.
For each example, it is useful to ask three questions in order:
- What happens to \(a_n\) as \(n\) increases?
- What is the formula for \(S_N\)?
- Does \(S_N\) approach a finite number?
Worked Example: A Geometric Sequence and Series
Worked Example: A Geometric Sequence and Series
Classify the term sequence and associated series when \(a_n=(1/3)^{n-1}\), for \(n\geq 1\).
The first terms are \(a_1=1\), \(a_2=1/3\), and \(a_3=1/9\). Since the powers of \(1/3\) approach zero, the term sequence converges to zero:
Now calculate the partial sums. For a positive integer \(N\),
Multiplying by \(1/3\) gives
Subtract the second equation from the first. Every term except the first and last cancels:
Therefore,
Because \((1/3)^N\to 0\), the partial sums converge to \(3/2\). Hence the associated series converges, and its sum is \(3/2\).
Check: For \(N=3\), direct addition gives \(S_3=1+1/3+1/9=13/9\). The formula gives \((3/2)(1-1/27)=(3/2)(26/27)=13/9\), as required.
Classification: The term sequence converges to \(0\); the partial-sum sequence converges to \(3/2\); the series converges to \(3/2\).
Why the Geometric Partial-Sum Formula Works
The cancellation above is a derivation, not just a convenient formula. In a finite geometric sum, multiplying by the common ratio shifts each term forward by one place. Subtracting the shifted sum removes all the shared middle terms. Only the first term and the final shifted term remain. Since the shifted final term approaches zero in this example, the finite partial sums approach a finite limit.
This reasoning also illustrates a general classification method: obtain an exact expression for \(S_N\), then take its limit. A conclusion about the series follows from the behavior of that partial-sum sequence.
Worked Example: A Telescoping Series
Worked Example: A Telescoping Series
Let \(a_n=\frac{1}{n(n+1)}\). Classify the terms, partial sums, and series.
First rewrite each term using partial fractions:
because \(\frac{1}{n}-\frac{1}{n+1}=\frac{(n+1)-n}{n(n+1)}=\frac{1}{n(n+1)}\). As \(n\to\infty\), the denominator \(n(n+1)\) increases without bound, so \(a_n\to 0\).
Using the rewritten term, the \(N\)th partial sum is
All intermediate terms cancel in pairs. Since \(1/(N+1)\to 0\), it follows that \(S_N\to 1\). Therefore the series converges to \(1\).
Check: For \(N=3\), direct addition gives \(S_3=1/2+1/6+1/12=6/12+2/12+1/12=9/12=3/4\). The formula gives \(1-1/4=3/4\). Also, the term formula is confirmed by \(1/3-1/4=1/12=1/[3(4)]\).
Classification: The term sequence converges to \(0\); the partial-sum sequence converges to \(1\); the series converges to \(1\).
A Convergent Term Sequence Does Not Settle the Series
The previous examples both have terms approaching zero and convergent series. That pattern in two examples does not mean that the behavior of the terms determines the behavior of the series. The definitions show that convergence of a series depends on the partial sums.
Worked Example: Constant Terms and Growing Partial Sums
Let \(a_n=2\) for every positive integer \(n\). Classify all three objects.
The term sequence is constant, so \(a_n=2\) for every \(n\), and therefore \(\lim_{n\to\infty}a_n=2\). Its partial sums are
For instance, \(S_1=2\), \(S_2=4\), and \(S_3=6\). As \(N\) increases, \(2N\) increases without bound. In particular, for any real number \(M\), choosing a positive integer \(N>M/2\) gives \(S_N=2N>M\). Thus the partial sums do not converge to a finite real number, and the series diverges.
Check: Substituting \(N=3\) into \(S_N=2N\) gives \(S_3=6\), which agrees with \(2+2+2=6\).
Classification: The term sequence converges to \(2\); the partial-sum sequence grows without bound; the series diverges.
This example is a useful reminder to avoid treating “the sequence” as if it named only one object. Here, the terms settle at \(2\), while the partial sums grow indefinitely.
Worked Example: Alternating Terms and Oscillating Partial Sums
Worked Example: Alternating Terms and Oscillating Partial Sums
Let \(a_n=(-1)^n\), starting at \(n=1\). Classify the term sequence, partial sums, and series.
The terms begin \(-1,1,-1,1,\ldots\). The odd-indexed terms are \(-1\), while the even-indexed terms are \(1\). These two subsequences have different limits, so the term sequence does not converge.
For the partial sums, pair consecutive terms. Each pair is \((-1)+1=0\). If \(N\) is even, all terms can be paired, so \(S_N=0\). If \(N\) is odd, there is one extra \(-1\), so \(S_N=-1\). Thus
The partial sums do not converge: along even indices they equal \(0\), while along odd indices they equal \(-1\). Since the partial-sum sequence has no finite limit, the series diverges.
Check: Directly, \(S_1=-1\), \(S_2=-1+1=0\), \(S_3=-1+1-1=-1\), and \(S_4=-1+1-1+1=0\). These values match the stated formula.
Classification: The term sequence does not converge; the partial-sum sequence does not converge; the series diverges.
A Reliable Classification Routine
Use the following steps to keep the objects and conclusions distinct:
Write down the formula for \(a_n\), and classify the sequence \(a_1,a_2,\ldots\) using its limit or its failure to have a limit.
Set \(S_N=\sum_{n=1}^{N}a_n\). Look for a finite-sum formula, cancellation, or a direct expression.
Determine whether \(S_N\) approaches a finite value, grows without bound, or fails to settle.
By definition, the series converges exactly when \(S_N\) converges to a finite real number. If it does, its sum is that limit.
Common Mistake: Confusing Terms with Partial Sums
A common incorrect argument is: “The sequence \(a_n=2\) converges, so the series \(\sum_{n=1}^{\infty}2\) converges.” This treats convergence of the terms as if it were convergence of the partial sums.
The correction is to calculate the partial sums directly:
Since \(2N\) grows without bound, the partial sums have no finite limit. Therefore the series diverges, even though the term sequence converges. Always identify whether a question asks about \(a_n\), \(S_N\), or \(\sum a_n\), and use the definition appropriate to that object.
Check Your Understanding
For each question, distinguish the sequence of terms from its partial sums before stating whether the series converges.
- For \(a_n=(1/2)^{n-1}\), what is the limit of \(a_n\), and what is the limit of \(S_N\)?
- For \(a_n=1/[n(n+1)]\), write a formula for \(S_N\) and state the series sum.
- If \(a_n=3\) for every \(n\), classify the term sequence and the associated series separately.
- For \(a_n=(-1)^{n+1}\), what values do the even and odd partial sums take? Does the series converge?
- In the definition of a convergent series, which sequence must have a finite limit?
