Infinite Sequences

Definitions

A sequence of real numbers is a function f (n), whose domain is the set of positive integers. The values a_n = f (n) taken by the function are called the terms of the sequence.

The set of values a_n = f (n) is denoted by {a_n}.

A sequence {a_n} has the limit L if for every ε > 0 there exists an integer N > 0 such that if n ≥ N, then |a_n − L| ≤ ε. In this case we write:

\[\lim\limits_{n \to \infty } {a_n} = L.\]

The sequence {a_n} has the limit ∞ if for every positive number M there is an integer N > 0 such that if n ≥ N then {a_n} > M. In this case we write

\[\lim\limits_{n \to \infty } {a_n} = \infty.\]

If the limit \(\lim\limits_{n \to \infty } {a_n} = L\) exists and L is finite, we say that the sequence converges. Otherwise the sequence diverges.

Squeezing Theorem

Suppose that \(\lim\limits_{n \to \infty } {a_n} = \lim\limits_{n \to \infty } {b_n} = L\) and \(\left\{ {{c_n}} \right\}\) is a sequence such that \({a_n} \le {c_n} \le {b_n}\) for all \(n \gt N,\) where \(N\) is a positive integer. Then

\[\lim\limits_{n \to \infty } {c_n} = L.\]

The sequence \(\left\{ {{a_n}} \right\}\) is bounded if there is a number \(M \gt 0\) such that \(\left| {{a_n}} \right| \le M\) for every positive \(n.\)

Every convergent sequence is bounded. Every unbounded sequence is divergent.

The sequence \(\left\{ {{a_n}} \right\}\) is monotone increasing if \({a_n} \le {a_{n + 1}}\) for every \(n \ge 1.\) Similarly, the sequence \(\left\{ {{a_n}} \right\}\) is called monotone decreasing if \({a_n} \ge {a_{n + 1}}\) for every \(n \ge 1.\) The sequence \(\left\{ {{a_n}} \right\}\) is called monotonic if it is either monotone increasing or monotone decreasing.

Solved Problems

Example 1.

Write a formula for the \(n\)th term of \({a_n}\) of the sequence and determine its limit (if it exists).

\[\frac{1}{3},\frac{2}{4},\frac{3}{5},\frac{4}{6},\frac{5}{7}, \ldots \]

Solution.

Here \({a_n} = {\frac{n}{{n + 2}}}.\) Then the limit is

\[\lim\limits_{n \to \infty } \frac{n}{{n + 2}} = \lim\limits_{n \to \infty } \frac{{n + 2 - 2}}{{n + 2}} = \lim\limits_{n \to \infty } \left( {1 - \frac{2}{{n + 2}}} \right) = \lim\limits_{n \to \infty } 1 - \lim\limits_{n \to \infty } \frac{2}{{n + 2}} = 1 - 0 = 1.\]

Thus, the sequence converges to \(1.\)

Example 2.

Write a formula for the \(n\)th term of \({a_n}\) of the sequence and determine its limit (if it exists).

\[1, - \frac{2}{2},\frac{3}{4}, - \frac{4}{8},\frac{5}{{16}}, \ldots \]

Solution.

We easily can see that \(n\)th term of the sequence is given by the formula \({a_n} = {\frac{{{{\left( { - 1} \right)}^{n - 1}}n}}{{{2^{n - 1}}}}}.\) Since \( - n \le {\left( { - 1} \right)^{n - 1}}n \le n,\) we can write:

\[\frac{{ - n}}{{{2^{n - 1}}}} \le \frac{{{{\left( { - 1} \right)}^{n - 1}}n}}{{{2^{n - 1}}}} \le \frac{n}{{{2^{n - 1}}}}.\]

Using L'Hopital's rule, we obtain

\[\lim\limits_{x \to \infty } \left( { \pm \frac{x}{{{2^{x - 1}}}}} \right) = \pm \lim\limits_{x \to \infty } \frac{x}{{{2^{x - 1}}}} = \pm \lim\limits_{x \to \infty } \frac{1}{{{2^{x - 1}}\ln 2}} = 0.\]

Hence, by the squeezing theorem, the limit of the initial sequence is

\[\lim\limits_{n \to \infty } \frac{{{{\left( { - 1} \right)}^{n - 1}}n}}{{{2^{n - 1}}}} = 0.\]

See more problems on Page 2.

Calculus