Calculus

Infinite Sequences and Series

Infinite Sequences and Series Logo

Infinite Sequences

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

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

    A sequence \(\left\{ {{a_n}} \right\}\) has the limit \(L\) if for every \(\varepsilon \gt 0\) there exists an integer \(N \gt 0\) such that if \(n \ge N,\) then \(\left| {{a_n} – L} \right| \le \varepsilon .\) In this case we write:

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

    The sequence \(\left\{ {{a_n}} \right\}\) has the limit \(\infty\) if for every positive number \(M\) there is an integer \(N \gt 0\) such that if \(n \ge N\) then \({a_n} \gt M.\) 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

    Click a problem to see the solution.

    Example 1

    Write a formula for the \(n\)th term of \({a_n}\) of the sequence and determine its limit (if it exists).
    \[\frac{1}{3},\frac{2}{4},\frac{3}{5},\frac{4}{6},\frac{5}{7}, \ldots \]

    Example 2

    Write a formula for the \(n\)th term of \({a_n}\) of the sequence and determine its limit (if it exists).
    \[1, – \frac{2}{2},\frac{3}{4}, – \frac{4}{8},\frac{5}{{16}}, \ldots \]

    Example 3

    Determine whether the sequence \(\left\{ {\large\frac{{2n + 3}}{{5n – 7}}\normalsize} \right\}\) converges or diverges.

    Example 4

    Does the sequence \(\left\{ {\large\frac{{{n^2}}}{{{2^n}}}\normalsize} \right\}\) converge or diverge?

    Example 5

    Determine whether the sequence \(\left\{ {\sqrt {n + 2} – \sqrt {n + 1} } \right\}\) converges or diverges.

    Example 6

    Determine whether the sequence \(\left\{ {\large\frac{{5n – 7}}{{3n + 4}}\normalsize} \right\}\) is increasing, decreasing, or neither.

    Example 7

    Determine whether the sequence \(\left\{ {\large\frac{{{2^n} + 3}}{{{2^n} + 1}}\normalsize} \right\}\) is increasing, decreasing, or not monotonic.

    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 \({a_n} = {\large\frac{n}{{n + 2}}\normalsize}.\) 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 \({a_n} = {\large\frac{{{{\left( { – 1} \right)}^{n – 1}}n}}{{{2^{n – 1}}}}\normalsize}.\) 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

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

    Page 1
    Problems 1-2
    Page 2
    Problems 3-7