When Small Terms Do Not Settle the Question
A series is an infinite sum, but its convergence is determined by a sequence of finite sums. This distinction matters because a sequence of terms can approach zero even while adding those terms produces partial sums that grow without bound.
We will compare two familiar series. The terms of both approach zero. One series has a finite sum; the other diverges. The comparison shows why checking the terms alone cannot decide whether a series converges.
Terms and Partial Sums
For a sequence of terms \(a_1,a_2,a_3,\ldots\), the \(N\)th partial sum is the finite sum of its first \(N\) terms. The series converges if the sequence of partial sums approaches a finite real number as \(N\) increases. If the partial sums do not approach a finite number, the series diverges.
The terms \(a_n\) and partial sums \(S_N\) are different sequences. A series can have terms that become very small, but the accumulated effect of many terms may still be large. The geometric and harmonic series make this difference precise.
A Geometric Series with a Finite Sum
A geometric series has terms that are successive powers of a fixed ratio. For example, in \(\sum_{n=1}^{\infty}(1/2)^n\), each term is half the term before it. The terms approach zero, and in this case the partial sums approach a finite value.
Worked Example: Summing a Geometric Series
Find the sum of \(\sum_{n=1}^{\infty}(1/2)^n\) by examining its partial sums.
The first \(N\) terms are \(S_N=\frac12+\frac14+\cdots+\frac{1}{2^N}\). Multiply by \(1/2\): \(\frac12S_N=\frac14+\frac18+\cdots+\frac{1}{2^{N+1}}\). Subtract the second equation from the first. All matching middle terms cancel, leaving \(S_N-\frac12S_N=\frac12-\frac{1}{2^{N+1}}\). Therefore, \(\frac12S_N=\frac12-\frac{1}{2^{N+1}}\), so \(S_N=1-\frac{1}{2^N}\).
As \(N\) increases, \(1/2^N\) approaches zero. Thus \(\lim_{N\to\infty}S_N=1\), and the series converges to \(1\).
Check: The partial-sum formula gives \(S_1=1-\frac12=\frac12\), matching the first term. Also, \(S_N-S_{N-1}=(1-\frac{1}{2^N})-(1-\frac{1}{2^{N-1}})=\frac{1}{2^N}\), which recovers the \(N\)th term.
The cancellation in this calculation explains why the geometric partial sums are manageable. For a general ratio \(r\), the same subtraction works for the finite sum \(r+r^2+\cdots+r^N\), provided \(r\ne1\): multiplying by \(r\) shifts each term one place, and subtraction cancels all but the first and last terms.
Consequently, \(S_N=\frac{r(1-r^N)}{1-r}\). When \(0<r<1\), the powers \(r^N\) approach zero, so these partial sums approach the finite value \(\frac{r}{1-r}\). The worked example used \(r=1/2\), giving \(\frac{r}{1-r}=1\). This is a result about the partial sums, not just about the terms becoming small.
The Harmonic Series: Small Terms, Unbounded Sums
The harmonic series is \(\sum_{n=1}^{\infty}\frac1n\). Its terms approach zero: the denominators grow without bound, so \(1/n\) becomes arbitrarily small. Yet its partial sums do not approach a finite value.
To see why, group terms into blocks whose endpoints are powers of two. The first block after the initial term contains \(1/2\). The next block contains \(1/3+1/4\). The following block contains \(1/5+1/6+1/7+1/8\). Each successive block has twice as many terms as the previous block, while every term in that block is at least as large as the reciprocal of its largest denominator.
Why the Harmonic Series Diverges
Here is the full grouping argument. Let \(H_N=1+\frac12+\cdots+\frac1N\). Consider \(H_{2^m}\), where \(m\) is a positive integer. After the initial term \(1\), group the terms into blocks from \(2^{k-1}+1\) through \(2^k\), for \(k=1,\ldots,m\). Each such block contains \(2^{k-1}\) terms. Every denominator in the block is at most \(2^k\), so each term is at least \(1/2^k\). Therefore, the sum of the block is at least \(2^{k-1}(1/2^k)=1/2\).
As \(m\) increases, \(1+m/2\) grows without bound. Thus the partial sums \(H_{2^m}\) are unbounded. Since all harmonic-series terms are positive, the partial sums \(H_N\) increase as \(N\) increases. In particular, the full sequence of partial sums cannot approach a finite limit when it has an unbounded subsequence. The harmonic series diverges.
Worked Example: Applying the Block Bound
Use the grouping argument to establish a lower bound for \(H_8=1+\frac12+\cdots+\frac18\).
Here \(8=2^3\), so take \(m=3\). The first block, from \(2\) through \(2\), contributes \(1/2\). The second, from \(3\) through \(4\), contributes at least \(2(1/4)=1/2\). The third, from \(5\) through \(8\), contributes at least \(4(1/8)=1/2\). Including the initial term gives \(H_8\ge 1+1/2+1/2+1/2=5/2\).
For a direct check, \(H_8=1+\frac12+\frac13+\frac14+\frac15+\frac16+\frac17+\frac18=\frac{761}{280}\), and \(\frac{761}{280}-\frac52=\frac{61}{280}>0\). The computed partial sum satisfies the claimed bound.
Why “The Terms Go to Zero” Is Not Enough
For a convergent series, its terms must approach zero. In fact, if \(S_N\) approaches a finite value \(S\), then the \(N\)th term satisfies \(a_N=S_N-S_{N-1}\). Since both partial sums approach \(S\), their difference approaches \(S-S=0\). So \(a_N\) must approach zero.
But this condition goes only one way: convergence requires terms to approach zero; it does not follow that a series converges whenever its terms approach zero. The harmonic series is the counterexample. Its terms vanish, but its partial sums grow without bound.
Worked Example: Rejecting an Invalid Conclusion
A student argues, “The terms \(1/n\) approach zero, so \(\sum_{n=1}^{\infty}1/n\) converges.” Identify the error and give the correct conclusion.
The term sequence satisfies \(\lim_{n\to\infty}1/n=0\). This verifies a necessary condition for convergence, not a sufficient one. For the partial sums, the block bound gives \(H_{2^m}\ge 1+m/2\). For example, \(H_{2^6}\ge 1+6/2=4\), and these lower bounds continue to grow as \(m\) increases. Hence the partial sums are unbounded, so the harmonic series diverges.
Check: The conclusion follows from the behavior of the partial sums, as the definition of series convergence requires. The limit of the individual terms alone does not establish convergence.
A Useful Comparison
Both examples have positive terms that approach zero. Their partial sums behave differently because their terms shrink at different rates and accumulate differently.
| Series | Limit of terms | Behavior of partial sums | Conclusion |
|---|---|---|---|
| \(\sum_{n=1}^{\infty}(1/2)^n\) | Zero | Approach \(1\) | Converges |
| \(\sum_{n=1}^{\infty}1/n\) | Zero | Grow without bound | Diverges |
This contrast is useful whenever you are asked to classify a series. First, distinguish the terms from the partial sums. Then remember that a nonzero term limit can rule out convergence, but a zero term limit cannot by itself prove convergence. Further tests can help classify other series, but the decisive object in the definition remains the sequence of partial sums.
Check Your Understanding
Answer each question using the distinction between terms and partial sums.
- What does it mean for \(\sum_{n=1}^{\infty}a_n\) to converge?
- For \(\sum_{n=1}^{\infty}(1/2)^n\), what is the \(N\)th partial sum, and what value does it approach?
- In the harmonic-series proof, why does each block from \(2^{k-1}+1\) through \(2^k\) contribute at least \(1/2\)?
- Why does an unbounded subsequence of harmonic partial sums show that the series cannot converge to a finite value?
- Does \(a_n\to0\) guarantee that \(\sum a_n\) converges? Give the example that supports your answer.
