Tutorials › Calculus II › The Harmonic Series as a Vanishing-Term Counterexample

Home  /  Tutorials  /  Basic Series Divergence

Basic Series Divergence · Tutorial 579 of 998

The Harmonic Series as a Vanishing-Term Counterexample

Use grouped partial sums to prove that the harmonic series diverges even though its terms approach zero.

Beginner 9 min read

What You'll Learn

  • Define convergence of a series through the limit of its partial sums.
  • Verify that the harmonic-series terms approach zero.
  • Group terms into blocks whose total contribution has a uniform lower bound.
  • Prove that selected partial sums grow without bound.
  • Distinguish a necessary condition for convergence from a sufficient one.
  • Identify the mistaken inference that a zero term limit guarantees convergence.

Why This Series Matters

A series can have terms that become arbitrarily small without its accumulated sum settling toward a finite value. The harmonic series is a basic, important example: its terms approach zero, but its partial sums grow without bound.

This distinction matters because the terms of a series and the sums of those terms answer different questions. A term describes one contribution. A partial sum records the total contribution so far. Even if each new contribution is small, many such contributions can accumulate to an arbitrarily large amount.

Series and Partial Sums

For a sequence of real numbers \(a_1,a_2,\ldots\), the \(N\)th partial sum is the finite sum of its first \(N\) terms. The infinite series is said to converge when the sequence of partial sums approaches a finite real number.

$$ S_N=\sum_{n=1}^{N}a_n, \qquad \sum_{n=1}^{\infty}a_n \text{ converges if } \lim_{N\to\infty}S_N \text{ exists as a finite real number.} $$

If the partial sums do not approach a finite real number, the series diverges. In this tutorial, we will show something stronger than mere failure to converge: the harmonic series’ partial sums are unbounded above.

Definition: The harmonic series is the series whose \(n\)th term is \(1/n\):
$$ \sum_{n=1}^{\infty}\frac{1}{n} = 1+\frac12+\frac13+\frac14+\cdots. $$
Its \(N\)th partial sum is denoted by \(H_N=\sum_{n=1}^{N}1/n\).

The Terms Do Approach Zero

Before examining the partial sums, verify the behavior of the individual terms. Since \(1/n\) is positive and decreases as \(n\) increases, it is reasonable to expect that it approaches zero. The definition of a sequence limit confirms this.

Let \(\varepsilon>0\). Choose a natural number \(N\) large enough that \(N>1/\varepsilon\). If \(n\geq N\), then \(n\) is positive and

$$ 0<\frac1n\leq\frac1N<\varepsilon. $$

Thus, for every positive tolerance \(\varepsilon\), all sufficiently late terms lie between zero and \(\varepsilon\). Therefore \(\lim_{n\to\infty}1/n=0\).

Important distinction: The limit of the terms describes the size of each contribution far out in the series. Convergence of the series concerns the limit of the accumulated partial sums. These are related questions, but they are not the same question.

Worked Example: Verify the Term Limit

Worked Example: Verify the Term Limit

Show directly that the terms \(a_n=1/n\) approach zero, and check the estimate for \(\varepsilon=0.01\).

For an arbitrary \(\varepsilon>0\), choose a natural number \(N>1/\varepsilon\). Whenever \(n\geq N\), the reciprocal inequality gives

$$ 0<\frac1n\leq\frac1N<\varepsilon. $$

This proves the limit by its definition. For the specific tolerance \(0.01=1/100\), choose \(N=101\), since \(101>100\). Then for every \(n\geq101\),

$$ 0<\frac1n\leq\frac1{101}<\frac1{100}=0.01. $$

The check is exactly the required inequality: every term from \(n=101\) onward differs from zero by less than \(0.01\). The general argument works for every positive \(\varepsilon\), not just this chosen tolerance.

Grouping the Partial Sums

To study accumulation, divide the terms into blocks whose endpoints are powers of two. Start with the first term, then group terms from \(2\) to \(2\), from \(3\) to \(4\), from \(5\) to \(8\), and continue. In general, the \(k\)th block after the first term runs from \(2^{k-1}+1\) through \(2^k\).

There are exactly \(2^{k-1}\) terms in that block. Each is at least \(1/2^k\), because its denominator is no larger than \(2^k\). Consequently, the whole block contributes at least

$$ 2^{k-1}\cdot\frac{1}{2^k}=\frac12. $$

This estimate is the key. The individual terms in later blocks are smaller than the earlier terms, but each block still contains enough terms to contribute at least one-half.

A Lower Bound for Selected Partial Sums

For a positive integer \(m\), group the terms in \(H_{2^m}\) into the first term and the \(m\) blocks just described. The first term contributes \(1\), and each block contributes at least \(1/2\). Therefore,

$$ H_{2^m} = 1+\sum_{k=1}^{m} \sum_{n=2^{k-1}+1}^{2^k}\frac1n \geq 1+\sum_{k=1}^{m}\frac12 = 1+\frac{m}{2}. $$

Every sum in this argument is finite, so grouping its terms does not change its value. The inequality follows by comparing each term in each block with \(1/2^k\), then adding those lower bounds.

Theorem: For every positive integer \(m\), the harmonic partial sum satisfies \(H_{2^m}\geq 1+m/2\). In particular, the harmonic partial sums are unbounded above, so the harmonic series diverges.

Why This Proves Divergence

To prove unboundedness, take any positive real threshold \(M\). Choose a positive integer \(m\) large enough that \(m>2(M-1)\). Such an integer exists because the positive integers have no upper bound. The lower estimate then gives

$$ H_{2^m}\geq 1+\frac{m}{2}>M. $$

Thus, no matter how large a threshold is chosen, some partial sum exceeds it. This is precisely what it means for the partial sums to be unbounded above.

A convergent sequence of partial sums must be bounded: if \(S_N\) tends to a finite limit \(L\), then all sufficiently late \(S_N\) lie within, for example, one unit of \(L\), and the remaining initial collection is finite and bounded as well. Since the harmonic partial sums are unbounded, they cannot converge to a finite limit. Therefore the harmonic series diverges.

1
Choose blocks.
Use the terms between consecutive powers of two.
2
Bound each block below.
The block from \(2^{k-1}+1\) through \(2^k\) has \(2^{k-1}\) terms, each at least \(1/2^k\), so its sum is at least \(1/2\).
3
Accumulate the bounds.
The partial sum through \(2^m\) is at least \(1+m/2\), which grows without bound as \(m\) increases.
4
Conclude divergence.
Unbounded partial sums cannot converge to a finite real number.

Worked Example: Bound the Partial Sum \(H_8\)

Worked Example: Bound the Partial Sum \(H_8\)

Group the terms of \(H_8\) into the first term and the blocks from \(2\) to \(2\), \(3\) to \(4\), and \(5\) to \(8\):

$$ H_8 = 1+\frac12 +\left(\frac13+\frac14\right) +\left(\frac15+\frac16+\frac17+\frac18\right). $$

The first block after \(1\) contributes exactly \(1/2\). In the block from \(3\) to \(4\), each term is at least \(1/4\), and there are two terms, so

$$ \frac13+\frac14\geq\frac14+\frac14=\frac12. $$

In the block from \(5\) to \(8\), each term is at least \(1/8\), and there are four terms, so

$$ \frac15+\frac16+\frac17+\frac18 \geq \frac18+\frac18+\frac18+\frac18 = \frac12. $$

Adding the first term and the three block lower bounds gives

$$ H_8\geq 1+\frac12+\frac12+\frac12=\frac52. $$

The check is that \(8=2^3\), so the general estimate gives \(H_8\geq1+3/2=5/2\), exactly the bound obtained by calculating the blocks.

Worked Example: Exceed Any Chosen Threshold

Worked Example: Exceed Any Chosen Threshold

Use the grouped-sum estimate to show that some harmonic partial sum exceeds \(10\).

We need \(1+m/2>10\). Subtracting \(1\) gives \(m/2>9\), so it is enough to choose an integer \(m>18\). Take \(m=19\). The corresponding partial sum ends at \(2^{19}=524288\), and

$$ H_{524288}=H_{2^{19}} \geq 1+\frac{19}{2} =\frac{21}{2} =10.5>10. $$

The check is direct: the lower bound \(10.5\) is strictly greater than the chosen threshold \(10\), so this partial sum, and hence at least one partial sum of the series, exceeds \(10\). The same choice process works for any positive threshold, establishing unboundedness rather than just one large value.

The General Lesson and a Common Mistake

The harmonic series demonstrates why a term limit of zero is not enough to prove convergence. A condition can be necessary without being sufficient. For a convergent series, its terms must approach zero; however, the harmonic series has terms approaching zero and still diverges.

Common Mistake: Treating a Necessary Condition as Sufficient

A mistaken argument might say:

$$ \lim_{n\to\infty}\frac1n=0, \quad\text{therefore}\quad \sum_{n=1}^{\infty}\frac1n\text{ converges}. $$

The first statement is correct, but the conclusion does not follow. It confuses a limit of individual terms with a limit of partial sums. The grouped lower bound instead shows that

$$ H_{2^m}\geq 1+\frac{m}{2}, $$

and the right-hand side grows without bound. Therefore the partial sums do not approach a finite value, and the series diverges. The correction is to test the accumulated partial sums, not to infer series convergence from the terms’ limit alone.

Grouping also reveals the mechanism behind this counterexample. In each successive block, the terms are individually smaller, but the number of terms in the block doubles. Those two effects balance sufficiently to keep each block’s total at least \(1/2\). Repeating such contributions without end forces the partial sums upward without bound.

Check Your Understanding

Use the definitions and estimates in this tutorial to answer the following questions.

  1. What does it mean for an infinite series to converge, in terms of its partial sums?
  2. How many terms are in the block from \(2^{k-1}+1\) through \(2^k\)?
  3. Why is every term in that block at least \(1/2^k\)?
  4. Use the block estimate to give a lower bound for \(H_{32}\).
  5. Why does \(\lim_{n\to\infty}1/n=0\) not prove that the harmonic series converges?

Previous

Next

Back to Tutorials