Sequence Behavior and Limiting Values

Definition Let $a_1, a_2, a_3, \dots$ be a sequence of real numbers and let $L$ be a real number. The sequence has limit $L$ if for every tolerance $\varepsilon > 0$ there is an integer $N$ such that

We then write $a_n \to L$ or

The inequality says that $a_n$ lies in the horizontal band between $L - \varepsilon$ and $L + \varepsilon$. So for every band around $L$, however thin, all terms from some index on lie inside it. Finitely many early terms may lie outside. Some books write $n \ge N$ in place of $n > N$, which only moves $N$ by one and gives the same limits.

Definition A sequence converges if it has a limit $L$ that is a real number, and then it converges to $L$. A convergent sequence has exactly one limit. Whether a sequence converges, and to what, depends only on its tail, the terms from some index on. Changing, adding or removing finitely many terms changes neither.

Definition A sequence diverges if it does not converge. One kind of divergence is growth without bound. We write $\lim_{n \to \infty} a_n = \infty$ when for every number $M$ there is an integer $N$ such that $a_n > M$ for every $n > N$, and $\lim_{n \to \infty} a_n = -\infty$ when for every $M$ the terms satisfy $a_n < M$ from some index on. Such a sequence still diverges, since $\infty$ is not a real number. The notation only says how it fails to converge.

Definition A sequence oscillates if it diverges but does not tend to $\infty$ or to $-\infty$. Its terms keep moving back and forth without settling. A simple test proves this kind of divergence. If there are numbers $p < q$ such that $a_n \le p$ for infinitely many $n$ and $a_n \ge q$ for infinitely many $n$, then the sequence diverges. The reason is that a limit would need a band of width $q - p$ around it to hold all terms from some index on. Two numbers inside such a band differ by less than $q - p$, while the tail always contains a term at most $p$ and a term at least $q$. Such a sequence cannot tend to $\infty$ or $-\infty$ either, since infinitely many terms stay at most $p$ and infinitely many stay at least $q$, so it oscillates.

Definition A sequence is bounded if there is a number $M$ with $\lvert a_n \rvert \le M$ for every $n$. Every convergent sequence is bounded. If $a_n \to L$, then all terms after some index $N$ lie between $L - 1$ and $L + 1$, and the finitely many terms up to $a_N$ have a largest absolute value. The converse is false, because a bounded sequence can diverge. What is true is that a bounded sequence that never decreases, or one that never increases, converges. That fact rests on the completeness of the real numbers and is proved in calculus.

Definition For a real number $r \ne 0$ the geometric sequence $r^n$, with $n = 0, 1, 2, \dots$, behaves as follows.

For $r \le -1$ the sequence oscillates. When $r = -1$ it alternates between $1$ and $-1$, and when $r < -1$ it alternates in sign while $\lvert r^n \rvert$ grows without bound. A sequence $c \, r^n$ with $c \ne 0$ has limit $0$ when $\lvert r \rvert < 1$ and limit $c$ when $r = 1$. It tends to $\infty$ or $-\infty$ when $r > 1$, depending on the sign of $c$, and oscillates when $r \le -1$. These facts decide when an infinite geometric series converges. For $r \ne 1$ and a first term $a \ne 0$ the sum of the first $n$ terms is

so the partial sums converge exactly when $r^n$ does, which means $\lvert r \rvert < 1$. For $r = 1$ the partial sums are $na$, which grow without bound in size. So with $a \ne 0$ the series converges exactly when $\lvert r \rvert < 1$, and then its sum is $\frac{a}{1 - r}$.

Definition Computing finitely many terms of a sequence, however many, does not prove that it converges, that it diverges, or what its limit is. The limit depends only on the tail, and every finite list stops at some index and leaves out all the terms after it. A proof is an argument that covers every index from some point on, such as an inequality showing $\lvert a_n - L \rvert < \varepsilon$ for all $n > N$ that works for every $\varepsilon > 0$.