Can a Convergent Sequence Be Unbounded?

If “sequence” means a sequence of real numbers with the usual definition of convergence, there is no convergent sequence that is unbounded. The reason is that convergence controls all sufficiently late terms, while only finitely many terms come before that point.

Proof for a real sequence

Suppose $(a_n)$ converges to $L$. By the definition of convergence, for every positive distance from $L$, all terms after some index lie within that distance. Choose the distance to be $1$. Then there is an integer $N$ such that

$$|a_n-L|<1 \qquad \text{whenever } n\ge N.$$

The triangle inequality gives, for every $n\ge N$,

$$|a_n|\le |a_n-L|+|L|<1+|L|.$$

Thus the tail of the sequence is bounded by $|L|+1$. The terms before $N$ form a finite list, so they are bounded as well; if there are no such terms, there is nothing to check. Taking a bound large enough to cover both the initial terms and $|L|+1$ gives one bound for every term of the sequence.

This proves that a convergent real sequence is bounded. The particular choice of distance $1$ is convenient but not essential; any fixed positive distance would work.

A concrete example

Consider

$$a_n=\frac{1}{n}, \qquad n=1,2,3,\ldots$$

This sequence converges to $0$. In fact, for every $n\ge1$,

$$0<\frac{1}{n}\le1,$$

so it is bounded by $1$. This illustrates the general result: the terms get close to the limit, and no finite collection of starting terms can make the whole sequence unbounded.

You might try $a_n=n$ as an unbounded sequence, but it does not converge to a real number. Its terms grow without bound, so it cannot approach a finite limit. A sequence can instead be bounded without converging: $a_n=(-1)^n$ stays between $-1$ and $1$ but keeps alternating between two values.

The same argument in a metric space

The result is not specific to real numbers. Suppose $(x_n)$ is a sequence in a metric space with distance function $d$, and $x_n$ converges to a point $x$. There is an index $N$ such that

$$d(x_n,x)<1 \qquad \text{for every } n\ge N.$$

The tail therefore lies within distance $1$ of $x$. The distances $d(x_n,x)$ for the finitely many earlier indices are bounded as well; if there are no such indices, there is nothing to check. Consequently, the set of terms of the sequence is bounded in the metric: there is some finite radius that contains every term.

For real numbers with the usual distance $d(a,b)=|a-b|$, this is exactly the proof above. In more general settings, “bounded” needs a specified notion of distance or size; convergence by itself does not define boundedness in a setting with no such notion.

Common mistakes

• Checking only the tail: Convergence gives a bound for all terms after some index, not necessarily the first few. You must also account for those finitely many initial terms.
• Assuming bounded means convergent: The implication goes one way: convergence implies boundedness. Boundedness alone does not guarantee convergence, as $(-1)^n$ shows.
• Confusing a sequence with its limit: A sequence may have a large finite term, but that does not make it unbounded. Unboundedness requires terms to exceed every proposed finite bound.
• Changing what convergence means: For an ordinary sequence of real numbers, the answer is always no. For sequences of functions or other objects, specify the mode of convergence and the notion of boundedness; those may not match the real-sequence setting.

So, under the standard interpretation, the requested example cannot exist. If a sequence of real numbers is unbounded, it cannot converge to a finite real limit.