Divergence of a Series

Divergence Is Failure of a Finite Limit

For a series \(\sum_{n=1}^{\infty}a_n\), write \(S_N=\sum_{n=1}^{N}a_n\) for its \(N\)th partial sum. The series converges when the sequence \((S_N)\) approaches a finite real number. To study divergence is therefore to understand the different ways in which the partial sums can fail to have such a limit.

Definition: The series \(\sum_{n=1}^{\infty}a_n\) diverges if its sequence of partial sums \((S_N)\) does not converge to any finite real number. If \(S_N\to+\infty\), the series is said to diverge to \(+\infty\); if \(S_N\to-\infty\), it is said to diverge to \(-\infty\). These are particular forms of divergence, not convergence to a real sum.

The definition makes the partial sums, rather than the terms considered one at a time, the central object. A sequence of partial sums may grow without bound, may oscillate, or may be bounded without settling toward a single value. Each behavior prevents convergence to a finite real number.

There is a useful way to express what the Cauchy Criterion for Series says about divergence. Since that criterion characterizes convergence by smallness of every sufficiently late finite block, its negation says that for a divergent series there is some fixed positive tolerance \(\varepsilon_0\) such that, no matter how far out one starts, a finite block beyond that point has absolute sum at least \(\varepsilon_0\). In symbols, there is an \(\varepsilon_0>0\) such that for every integer \(N\), there are integers \(p>q\geq N\) with

\left|\sum_{n=q+1}^{p}a_n\right|\geq\varepsilon_0.

This formulation is often useful when partial sums do not have an easy explicit formula. It says that divergence must leave some fixed-size change in arbitrarily late blocks. It does not say that every late block is large: the condition requires only that at least one such block can be found after each starting point.

Unbounded Partial Sums Force Divergence

A convergent sequence of real numbers is bounded. Applied to partial sums, this familiar fact immediately gives a sufficient condition for divergence: if the partial sums are unbounded, the series cannot converge. The following proof includes the boundedness argument so that the role of the finite initial segment is explicit.

Theorem (Unbounded Partial Sums Imply Divergence): Let \(S_N=\sum_{n=1}^{N}a_n\). If the sequence \((S_N)\) is unbounded, then \(\sum_{n=1}^{\infty}a_n\) diverges.

Proof. Suppose, to the contrary, that the series converges to some \(S\in\mathbb{R}\). Then \(S_N\to S\). By the definition of sequence convergence, there is an integer \(N_0\) such that \(N\geq N_0\) implies \(|S_N-S|<1\). The triangle inequality then gives

|S_N|\leq |S|+|S_N-S|<|S|+1 \qquad (N\geq N_0).

The finitely many values \(S_1,\ldots,S_{N_0-1}\), if there are any, are also bounded. For example, take \(B\) to be the larger of \(|S|+1\) and the maximum of those finitely many absolute values; if \(N_0=1\), take \(B=|S|+1\). Then \(|S_N|\leq B\) for every \(N\), contradicting the assumed unboundedness. Therefore the series diverges. \(\square\)

This theorem is a one-way test. Unbounded partial sums prove divergence, but bounded partial sums alone do not prove convergence. A bounded sequence may keep moving among different values. To handle that possibility, it is useful to look at subsequences of partial sums.

Worked Example: The Harmonic Series Diverges to Infinity

Consider \(\sum_{n=1}^{\infty}1/n\). Its terms are positive, so its partial sums increase. We show that the partial sums are unbounded by grouping terms between successive powers of \(2\). For each integer \(j\geq1\), consider the indices

2^{j-1}+1\leq n\leq 2^j.

There are \(2^{j-1}\) integers in this block. Each is at most \(2^j\), so each corresponding term satisfies \(1/n\geq1/2^j\). Consequently,

\sum_{n=2^{j-1}+1}^{2^j}\frac1n \geq 2^{j-1}\cdot\frac1{2^j} = \frac12.

For an integer \(m\geq1\), the first \(m\) such blocks, together with the first term, give

S_{2^m} = 1+\sum_{j=1}^{m}\sum_{n=2^{j-1}+1}^{2^j}\frac1n \geq 1+\frac{m}{2}.

For \(m=1\), this estimates \(S_2=1+1/2\) by \(1+1/2\). For \(m=2\), the next block contributes \(1/3+1/4\geq1/2\), so \(S_4\geq2\). As \(m\) increases, the lower bound \(1+m/2\) grows without bound. Thus the partial sums are unbounded, and the Unbounded Partial Sums Imply Divergence theorem proves that the harmonic series diverges. In fact, because its partial sums are increasing and unbounded, they tend to \(+\infty\).

Different Subsequence Limits Show Divergence

Sometimes the partial sums remain bounded, so unboundedness is not available as a test. Another decisive sign of divergence is that two subsequences of partial sums approach different limits. If the full sequence had a limit, every subsequence would have to approach that same limit.

Theorem (Distinct Subsequence Limits Imply Divergence): Let \(S_N=\sum_{n=1}^{N}a_n\). Suppose there are two strictly increasing sequences of indices \((m_k)\) and \((n_k)\) such that \(S_{m_k}\to L\) and \(S_{n_k}\to M\), where \(L\neq M\). Then \(\sum_{n=1}^{\infty}a_n\) diverges.

Proof. Suppose instead that the series converges to \(S\in\mathbb{R}\), so \(S_N\to S\). Every subsequence of a convergent sequence converges to the same limit: indeed, for any \(\varepsilon>0\), once all indices \(N\) beyond some \(N_0\) satisfy \(|S_N-S|<\varepsilon\), the same inequality holds for every subsequence term whose index is at least \(N_0\). Thus both \(S_{m_k}\) and \(S_{n_k}\) must converge to \(S\). Limits of real sequences are unique, so \(L=S=M\), contradicting \(L\neq M\). Therefore the series diverges. \(\square\)

Worked Example: A Bounded Sequence of Partial Sums That Oscillates

Consider the series \(\sum_{n=1}^{\infty}(-1)^{n+1}\), whose terms are \(1,-1,1,-1,\ldots\). Pairing consecutive terms gives \(1+(-1)=0\), and direct calculation yields

S_{2k}=0 \qquad\text{and}\qquad S_{2k-1}=1 \quad (k\geq1).

For example, \(S_1=1\), \(S_2=1-1=0\), \(S_3=1-1+1=1\), and \(S_4=1-1+1-1=0\), in agreement with the formulas. The even-indexed partial sums form a subsequence converging to \(0\), while the odd-indexed partial sums form a subsequence converging to \(1\). Since these limits are distinct, the Distinct Subsequence Limits Imply Divergence theorem proves that the series diverges. Its partial sums are bounded between \(0\) and \(1\), so this example also shows why boundedness alone is not a convergence test.

Worked Example: Sparse Nonzero Terms Still Give Unbounded Partial Sums

Define \(a_n=1\) when \(n\) is a power of \(2\), and \(a_n=0\) otherwise. Among the indices from \(1\) through \(N\), the powers of \(2\) are \(2^0,2^1,\ldots,2^{\lfloor\log_2 N\rfloor}\). Hence

S_N=\lfloor\log_2 N\rfloor+1.

The formula can be checked at the first few indices: \(S_1=1\), because \(1=2^0\) is the only power of \(2\) not exceeding \(1\); \(S_2=2\), because \(1\) and \(2\) are powers of \(2\); and \(S_4=3\), because the powers of \(2\) not exceeding \(4\) are \(1,2,4\). In particular, the nonzero terms through index \(4\) contribute \(1+1+1=3\). As \(N\) increases, the number of powers of \(2\) not exceeding \(N\) is unbounded. Therefore \((S_N)\) is unbounded, and the series diverges. The gaps between the nonzero terms grow, but the accumulated partial sums still increase without bound.

Choosing a Divergence Argument

The examples illustrate two different strategies. For positive terms, grouping can show that the partial sums grow beyond every fixed bound. When signs vary, subsequences can reveal persistent oscillation even if the partial sums stay bounded. The right argument depends on the behavior of the partial sums, not merely on the appearance of a few terms.

Write the partial sums.
Begin with \(S_N=\sum_{n=1}^{N}a_n\). If a formula or useful grouping is available, use it to examine the full sequence of sums.

Check for unboundedness.
If the partial sums can be shown to exceed arbitrarily large bounds in absolute value, apply the Unbounded Partial Sums Imply Divergence theorem.

Look for incompatible subsequences.
If the sums are bounded or oscillatory, find two subsequences with different limits. The Distinct Subsequence Limits Imply Divergence theorem then settles the question.

Use the Cauchy criterion when blocks are easier to estimate.
To prove divergence without explicit partial sums, show that finite blocks beginning arbitrarily far out fail to become uniformly small.

A common pitfall is to infer convergence from the fact that the terms are small or the partial sums are bounded. Neither observation establishes that all partial sums settle toward one finite value. The Cauchy Criterion for Series requires control of every sufficiently late finite block, while the subsequence test detects failure to settle by comparing limiting behavior along different index sequences.

The Term Test for Series Convergence, established earlier in “What Is an Infinite Series?”, states that convergence forces \(a_n\to0\). That result is useful in the other direction: if the terms do not tend to zero, the series cannot converge. But when terms do tend to zero, further analysis is needed. The harmonic series is one example: its terms approach zero, yet its partial sums grow without bound. The next tutorial examines the necessary condition for series convergence more closely.

Takeaway: A series diverges when its partial sums fail to approach a finite real limit. Unbounded partial sums give one direct proof; two subsequences with different limits give another. Bounded partial sums, by themselves, do not rule out divergence.

Check Your Understanding

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

What sequence determines whether the series \(\sum_{n=1}^{\infty}a_n\) converges or diverges?
Why do unbounded partial sums rule out convergence to a finite real number?
For the harmonic series, why does the block from \(2^{j-1}+1\) through \(2^j\) have sum at least \(1/2\)?
How can two subsequences of partial sums prove divergence even when all partial sums are bounded?
For the series whose terms are \(1\) at powers of \(2\) and \(0\) elsewhere, list the indices of the nonzero terms through \(N=8\) and calculate \(S_8\).
Why does boundedness of the partial sums not, on its own, establish convergence?

What You'll Learn