When Terms Become Small, What Can We Conclude?
A series adds the terms of a sequence in order. If those terms become small, it is tempting to expect that their total must settle to a finite value. But “small” terms can accumulate without bound. To decide whether a series converges, we must examine its partial sums, not just the size of its individual terms.
The earlier material on necessary conditions for series convergence established that the terms of a convergent series must approach zero. This tutorial focuses on the crucial limitation of that fact: a zero term limit does not prove convergence. We will see a convergent geometric series and divergent examples where the terms also approach zero.
Series Convergence and the Nth-Term Test
For a sequence of terms \(a_1,a_2,a_3,\ldots\), the \(N\)th partial sum is the sum of its first \(N\) terms. The infinite series is said to converge when the sequence of partial sums approaches a finite real number.
If the partial sums do not approach a finite real number, the series diverges. The limit of the terms, \(\lim_{n\to\infty}a_n\), is a different limit from the limit of the partial sums. Confusing these two limits is a common source of errors.
The test is often called the nth-term test for divergence. Its reliable conclusion is one-way: a nonzero term limit, or a term limit that does not exist, proves divergence. A zero term limit does not establish convergence.
Why Convergence Requires Terms to Approach Zero
The reason is a direct relationship between consecutive partial sums. For \(n\geq 2\), the \(n\)th term is the difference between the \(n\)th partial sum and the preceding partial sum. If both partial sums approach the same finite limit, their difference must approach zero.
Since \(S_n=a_1+\cdots+a_n\) and \(S_{n-1}=a_1+\cdots+a_{n-1}\), subtracting gives \(a_n=S_n-S_{n-1}\).
If \(S_n\to L\), then the shifted sequence \(S_{n-1}\) also approaches \(L\).
Both partial-sum sequences approach \(L\), so their difference approaches \(L-L=0\). Hence \(a_n\to0\).
In equations, the argument is
This proves that a zero term limit is necessary for convergence. It does not prove that the condition is sufficient. A necessary condition must hold whenever the desired conclusion holds; it need not guarantee that conclusion on its own.
A Convergent Example: A Geometric Series
Worked Example: A Geometric Series with Ratio One-Half
Consider \(\sum_{n=1}^{\infty}(1/2)^n\). Its first terms are \(1/2,1/4,1/8,\ldots\). We will find the limit of its partial sums, rather than infer convergence from the term limit alone.
The finite geometric-sum formula gives
As \(N\to\infty\), \((1/2)^N\to0\). Therefore
The series converges to \(1\). Its terms also approach zero, since \(\lim_{n\to\infty}(1/2)^n=0\). The conclusion follows from the partial sums; the term limit alone would not have been enough.
Check: The first three terms sum to \(1/2+1/4+1/8=7/8\). The formula gives \(S_3=1-(1/2)^3=7/8\), matching the direct calculation. Differencing the formula also recovers the terms: \(S_N-S_{N-1}=(1-(1/2)^N)-(1-(1/2)^{N-1})=(1/2)^N\).
A Divergent Example: Terms of the Form \(1/\sqrt{n}\)
Now consider the positive terms \(a_n=1/\sqrt{n}\). They do approach zero, but the partial sums do not approach a finite limit. Grouping terms into intervals whose sizes double makes the accumulation visible.
Worked Example: A Divergent Series with Terms \(1/\sqrt{n}\)
For each integer \(k\geq0\), group the terms with indices \(2^k\leq n\leq2^{k+1}-1\). There are \(2^k\) terms in this group. Each of them is greater than \(2^{-(k+1)/2}\), because \(n<2^{k+1}\) implies \(\sqrt n<2^{(k+1)/2}\). Thus the sum of the group is greater than
For every \(k\geq1\), \(2^{(k-1)/2}\geq1\). Consequently, each group with \(k\geq1\) contributes more than \(1\). The partial sum through the end of the group indexed by \(K\), where \(K\geq1\), therefore satisfies
These partial sums are unbounded as \(K\to\infty\). Since a convergent sequence of partial sums must be bounded, the series diverges. Meanwhile, \(1/\sqrt n\to0\): given any \(\varepsilon>0\), choosing \(n>1/\varepsilon^2\) ensures \(1/\sqrt n<\varepsilon\).
Check: For \(K=2\), the endpoint is \(2^3-1=7\). The groups are \(\{1\}\), \(\{2,3\}\), and \(\{4,5,6,7\}\); the proof’s bounds show \(S_7>3\). In general, the endpoint bound grows with \(K\), so the partial sums cannot settle to a finite limit.
A Second Divergent Example: Small Terms Collected in Long Blocks
A custom sequence makes the distinction between term size and accumulated sum especially clear. In each successive block, the terms become smaller, but the block contains enough terms to contribute a fixed amount.
Worked Example: Blocks with Total Contribution One
For each \(k\geq1\), assign the value \(2^{-k}\) to the \(2^k\) consecutive terms whose indices run from \(2^k-1\) through \(2^{k+1}-2\). The first blocks are: \(a_1=a_2=1/2\); \(a_3,\ldots,a_6=1/4\); and \(a_7,\ldots,a_{14}=1/8\). The block endpoints follow because the number of terms in all preceding blocks is \(2+4+\cdots+2^{k-1}=2^k-2\).
Within block \(k\), each term equals \(2^{-k}\), and there are \(2^k\) such terms. The block therefore contributes exactly
After the first \(K\) complete blocks, the partial sum is exactly \(K\). The endpoint index is \(2^{K+1}-2\), so
These partial sums grow without bound, proving that the series diverges. Yet its terms approach zero: as \(n\) increases, its block index \(k\) increases, and the assigned value \(2^{-k}\) tends to zero.
Check: The first block contributes \(2\cdot(1/2)=1\), and the second contributes \(4\cdot(1/4)=1\). Hence \(S_6=2\), as the formula predicts for \(K=2\). Each additional complete block adds another \(1\), so the partial sums cannot converge to a finite value.
The Common Mistake and Its Correction
A frequent incorrect argument is: “The terms of \(\sum 1/\sqrt n\) approach zero, so the series converges.” The premise is true, but the conclusion does not follow. The dyadic-group calculation shows why: even as individual terms shrink, the groups of terms accumulate more than \(1\) each from \(k=1\) onward.
The test’s outcomes can be summarized as follows:
| Limit of the terms | What the nth-term test establishes |
|---|---|
| \(\lim_{n\to\infty}a_n\neq0\) | The series diverges. |
| The term limit does not exist | The series diverges. |
| \(\lim_{n\to\infty}a_n=0\) | No conclusion about convergence or divergence. |
Why the Distinction Matters
The terms describe individual contributions; the partial sums describe the accumulated total. A sequence of contributions can shrink while their cumulative effect still grows. The geometric example converges because its partial sums approach \(1\). The other examples diverge because their partial sums become unbounded. All three have terms approaching zero, so that condition by itself cannot distinguish the outcomes.
When you encounter a series, the term limit is still useful. If it is nonzero or fails to exist, divergence is settled immediately. If it is zero, do not stop and declare convergence: that is precisely the case in which the nth-term test has no answer. A separate argument about the partial sums is required.
Check Your Understanding
Answer each question using the distinction between terms and partial sums.
- Why must \(a_n\to0\) if the partial sums \(S_n\) converge?
- If \(\lim_{n\to\infty}a_n=0\), what can the nth-term test conclude about \(\sum a_n\)?
- For \(\sum_{n=1}^{\infty}(1/2)^n\), what is \(S_N\), and what limit does it approach?
- In the block example, how many terms are in block \(k\), and what is their total contribution?
- Give one reason that terms approaching zero do not guarantee that a series converges.
