Calculus

Infinite Sequences and Series

Infinite Sequences and Series Logo

Infinite Series – Page 2

  • Example 3.

    Show that the harmonic series \(\sum\limits_{n = 1}^\infty {\large\frac{1}{n}\normalsize} \) diverges.

    Solution.

    To see this, we can write

    \[ {\sum\limits_{n = 1}^\infty {\frac{1}{n}} } = {1 + \frac{1}{2} + \underbrace {\left( {\frac{1}{3} + \frac{1}{4}} \right)}_{\frac{7}{{12}} > \frac{1}{2}} } + {\underbrace {\left( {\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}} \right)}_{\frac{{533}}{{840}} > \frac{1}{2}} }+{ \ldots \;\text{and so on}.} \]

    Therefore \(\sum\limits_{n = 1}^\infty {\large\frac{1}{n}\normalsize} \) \(> \sum\limits_{n = 1}^\infty {\large\frac{1}{2}\normalsize} \) = \(\infty .\) Hence, the harmonic series diverges.

    Actually, this result was first proved by a mediaeval French mathematician, Nichole Oresme, who lived over \(600\) years ago.

    Example 4.

    Investigate convergence of the series \(\sum\limits_{n = 0}^\infty {\left( {\large\frac{1}{{{3^n}}}\normalsize} + {\large\frac{1}{{{5^n}}}\normalsize} \right)}.\)

    Solution.

    This series converges because it is the sum of two convergent series, \(\sum\limits_{n = 0}^\infty {\large\frac{1}{{{3^n}}}\normalsize} \) and \(\sum\limits_{n = 0}^\infty {\large\frac{1}{{{5^n}}}\normalsize}.\) Both are geometric series with ratio \(\left| q \right| \lt 1.\) Then

    \[
    {\sum\limits_{n = 0}^\infty {\frac{1}{{{3^n}}}} }
    = {\sum\limits_{n = 0}^\infty {{{\left( {\frac{1}{3}} \right)}^n}} }
    = {\frac{1}{{1 – \frac{1}{3}}} }={ \frac{3}{2},}
    \]

    \[
    {\sum\limits_{n = 0}^\infty {\frac{1}{{{5^n}}}} }
    = {\sum\limits_{n = 0}^\infty {{{\left( {\frac{1}{5}} \right)}^n}} }
    = {\frac{1}{{1 – \frac{1}{5}}} }={ \frac{5}{4}.}
    \]

    Hence, the sum of the given series is

    \[
    {\sum\limits_{n = 0}^\infty {\left( {\frac{1}{{{3^n}}} + \frac{1}{{{5^n}}}} \right)} }
    = {\frac{3}{2} + \frac{5}{4} }={ \frac{{11}}{4}.}
    \]

    Example 5.

    Investigate convergence of the series \(\sum\limits_{n = 1}^\infty {\large\frac{1}{{\left( {n + \pi } \right)\left( {n + \pi + 1} \right)}}\normalsize}.\)

    Solution.

    We see that

    \[{\frac{1}{{\left( {n + \pi } \right)\left( {n + \pi + 1} \right)}} }={ \frac{1}{{n + \pi }} }-{ \frac{1}{{n + \pi + 1}}.}\]

    Then the \(n\)th partial sum is

    \[ {{S_n} = \left( {\frac{1}{{1 + \pi }} – \frac{1}{{2 + \pi }}} \right) } + {\left( {\frac{1}{{2 + \pi }} – \frac{1}{{3 + \pi }}} \right) + \ldots } + {\left( {\frac{1}{{n + \pi }} – \frac{1}{{n + \pi + 1}}} \right) } = {\frac{1}{{1 + \pi }} }-{ \frac{1}{{n + \pi + 1}}.} \]

    Calculate the limit of \({S_n}\) as \(n \to \infty:\)

    \[
    {\lim\limits_{n \to \infty } {S_n} }
    = {\lim\limits_{n \to \infty } \left( {\frac{1}{{1 + \pi }} – \frac{1}{{n + \pi + 1}}} \right) }
    = {\frac{1}{{1 + \pi }} }\approx{ 0,24.}
    \]

    Hence, the series converges.

    Example 6.

    Determine whether the series
    \[{\frac{1}{{1 \cdot 2}} + \frac{1}{{2 \cdot 3}} }+{ \frac{1}{{3 \cdot 4}} + \frac{1}{{4 \cdot 5}} + \ldots }+{ \frac{1}{{n\left( {n + 1} \right)}} + \ldots }\]
    converges or diverges.

    Solution.

    The \(n\)th partial sum is

    \[{{S_n} = \frac{1}{{1 \cdot 2}} + \frac{1}{{2 \cdot 3}} }+{ \frac{1}{{3 \cdot 4}} + \frac{1}{{4 \cdot 5}} + \ldots }+{ \frac{1}{{n\left( {n + 1} \right)}}.}\]

    We can easily see that

    \[
    {\frac{1}{{1 \cdot 2}} = 1 – \frac{1}{2},\;\;}\kern-0.3pt
    {\frac{1}{{2 \cdot 3}} = \frac{1}{2} – \frac{1}{3},\;\;}\kern-0.3pt
    {\frac{1}{{3 \cdot 4}} = \frac{1}{3} – \frac{1}{4},\;\;}\kern-0.3pt
    {\frac{1}{{4 \cdot 5}} = \frac{1}{4} – \frac{1}{5},\;\; \ldots,\;\;}\kern-0.3pt
    {{\frac{1}{{n\left( {n + 1} \right)}} }={ \frac{1}{n} – \frac{1}{{n + 1}}.}}
    \]

    Then

    \[ {{S_n} = \left( {1 – \frac{1}{2}} \right) + \left( {\frac{1}{2} – \frac{1}{3}} \right) } + {\left( {\frac{1}{3} – \frac{1}{4}} \right) + \ldots } + {\left( {\frac{1}{n} – \frac{1}{{n + 1}}} \right).} \]

    Hence,

    \[
    {{S_n} = 1 – \frac{1}{{n + 1}}\;\;\text{and}\;\;}\kern0pt
    {{\lim\limits_{n \to \infty } {S_n} }={ \lim\limits_{n \to \infty } \left( {1 – \frac{1}{{n + 1}}} \right) }={ 1.}}
    \]

    Thus the series converges to \(1.\)

    Example 7.

    Evaluate \(\sum\limits_{n = 0}^\infty {\ln {\large\frac{{n + 2}}{{n + 1}}\normalsize}}.\)

    Solution.

    We write the \(n\)th term as

    \[
    {{a_n} = \ln \frac{{n + 2}}{{n + 1}} }
    = {\ln \left( {n + 2} \right) }-{ \ln \left( {n + 1} \right).}
    \]

    Calculate the \(n\)th partial sum:

    \[ {{S_n} }={ \left( {\ln 2 – \ln 1} \right) }+{ \left( {\ln 3 – \ln 2} \right) } + {\left( {\ln 4 – \ln 3} \right) + \ldots } + {\left[ {\ln \left( {n + 2} \right) – \ln \left( {n + 1} \right)} \right] } = {\left( { – \ln 1 + \ln 2} \right) }+{ \left( { – \ln 2 + \ln 3} \right) } + {\left( { – \ln 3 + \ln 4} \right) + \ldots } + {\left[ { – \ln \left( {n + 1} \right) + \ln \left( {n + 2} \right)} \right] } = { – \ln 1 + \ln \left( {n + 2} \right) } = {\ln \left( {n + 2} \right).} \]

    Since \(\lim\limits_{n \to \infty } {S_n} \) \(= \lim\limits_{n \to \infty } \left[ {\ln \left( {n + 2} \right)} \right] \) \(= \infty ,\) we conclude that the given series diverges.

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