Sequences Diverging to Infinity

When Terms Eventually Pass Every Bound

A sequence can diverge in several ways. It may oscillate among bounded values, or its terms may become arbitrarily large in magnitude without settling in one direction. In this tutorial, we focus on a more specific behavior: the terms eventually rise above every fixed real bound, or eventually fall below every fixed real bound. This distinction gives a useful way to describe sequences whose terms move without bound in a consistent direction.

The word “eventually” is essential. A sequence tending to positive infinity need not be increasing at every step, and it need not be positive at every index. What matters is that, after a suitable index depending on the bound, every later term exceeds that bound. The threshold index may change when the bound changes.

These symbols describe behavior, not real-number limits: \(+\infty\) and \(-\infty\) are not real numbers. For \(a_n\to+\infty\), one may choose any real \(M\), including a large positive number or a negative number, and the sequence eventually stays above it. For \(a_n\to-\infty\), the sequence eventually stays below every real \(M\).

The quantifiers also clarify a common distinction. To tend to positive infinity, terms greater than \(M\) must occur at every sufficiently late index, not merely at arbitrarily late indices. The latter condition can show unboundedness, but it does not rule out repeated returns to small values.

Examples of Divergence to Infinity

Subsequences Preserve the Direction

A sequence tending to positive infinity cannot have a subsequence that repeatedly returns below some fixed bound. The same eventual inequalities that hold for the whole sequence must also hold along any increasing list of indices. The index estimate from earlier in the course, \(n_k\geq k\) for a strictly increasing sequence of nonnegative integer indices, makes this precise.

Proof. We prove the positive-infinity statement. Let \((a_{n_k})\) be any subsequence, so \(n_0<n_1<n_2<\cdots\). Fix \(M\in\mathbb{R}\). Since \(a_n\to+\infty\), there is an \(N\in\mathbb{N}_0\) such that \(a_n>M\) for every \(n\geq N\). For every \(k\geq N\), the index estimate gives \(n_k\geq k\geq N\), and therefore \(a_{n_k}>M\). This is exactly the definition of \(a_{n_k}\to+\infty\).

For the negative-infinity statement, fix \(M\in\mathbb{R}\). There is an \(N\) such that \(a_n<M\) for every \(n\geq N\). If \(k\geq N\), then \(n_k\geq k\geq N\), so \(a_{n_k}<M\). Thus \(a_{n_k}\to-\infty\). \(\square\)

In particular, a sequence tending to positive infinity cannot have a subsequence converging to a real number. Indeed, every subsequence also tends to positive infinity, whereas a sequence converging to a real number is eventually trapped in a bounded interval around that number. This is one way that divergence to infinity differs from bounded oscillation.

Comparison Gives a Direct Test

A lower bound can prove that a sequence tends to positive infinity if the lower bound itself tends to positive infinity. There is also a useful version that compares two sequences directly. The inequalities need only hold eventually: changing finitely many terms does not affect the conclusion, in keeping with the finite-change results from earlier in the course.

Proof. Assume \(a_n\leq b_n\) for every \(n\geq N_0\) and \(a_n\to+\infty\). Fix \(M\in\mathbb{R}\). By the definition of divergence to positive infinity, there is an \(N_1\in\mathbb{N}_0\) such that \(a_n>M\) for every \(n\geq N_1\). Set \(N=\max\{N_0,N_1\}\). For every \(n\geq N\), both \(a_n\leq b_n\) and \(a_n>M\) hold, so \(b_n\geq a_n>M\). Since \(M\) was arbitrary, \(b_n\to+\infty\).

For the negative-infinity assertion, suppose \(b_n\leq a_n\) eventually and \(a_n\to-\infty\). Fix \(M\in\mathbb{R}\). Choose \(N_1\) such that \(a_n<M\) for every \(n\geq N_1\), and let \(N\) be at least \(N_0\) and \(N_1\). For every \(n\geq N\), \(b_n\leq a_n<M\), so \(b_n<M\). This holds for every \(M\), proving \(b_n\to-\infty\). \(\square\)

Adding a Sequence That Is Bounded Below

The comparison example suggests a broader principle: adding a sequence that is bounded below cannot stop a sequence tending to positive infinity from doing so. The added sequence does not have to converge or be bounded above. A fixed lower bound is enough.

Proof. Fix \(M\in\mathbb{R}\). Since \(a_n\to+\infty\), there is an \(N\in\mathbb{N}_0\) such that \(a_n> M-B\) for every \(n\geq N\). For each such \(n\), the assumed lower bound gives

Thus \(a_n+b_n>M\) for every \(n\geq N\). As this works for every real \(M\), \(a_n+b_n\to+\infty\). \(\square\)

Divergence to Infinity Is Stronger Than Unboundedness

If \(a_n\to+\infty\), then the sequence is unbounded above: for any proposed upper bound \(U\), the definition with \(M=U\) gives terms—and in fact all sufficiently late terms—greater than \(U\). Similarly, \(a_n\to-\infty\) implies that the sequence is unbounded below. The earlier theorem Unbounded Sequences Diverge therefore shows that either kind of divergence to infinity implies divergence in the ordinary real-limit sense.

The converse is not true. Unboundedness alone does not say that all sufficiently late terms stay on one side of a bound; a sequence may have arbitrarily large terms in both directions. Even being unbounded above does not imply convergence to positive infinity if the sequence keeps returning below a fixed number. For instance, define \(u_n=n\) for even \(n\) and \(u_n=0\) for odd \(n\). The even-indexed terms grow without bound, but every odd-indexed term is zero, so the sequence does not tend to positive infinity. Its behavior at arbitrarily late indices, not just its largest values, determines whether it diverges to infinity.

When proving a claim of divergence to positive or negative infinity, check the final inequality for every \(n\) beyond one chosen threshold. Showing only that large or small terms occur infinitely often proves less. A comparison bound, as in the examples above, is often an efficient way to meet the full “eventually every term” requirement.