Convergent and Divergent Series

\( \) \( \) \( \) \( \)\( \) \( \) \( \) \( \)

Convergence and divergence of Series are presented with examples and their detailed solutions.


Definition of Series

Let \( \displaystyle \left\{a_n \right\}_1^{\infty} \) be an infinite sequence . The infinite sum \[ a_1 + a_2 + a_3 + ... + \; a_n \; + ... \] is called a series. [1] [2]
The series may also be written using the symbol \( \displaystyle \sum \) as \[ \displaystyle \sum_{i=1}^{\infty} a_i \]
Note that when the upper limit of the sum is infinite as in \( \displaystyle \sum_{i=1}^{\infty} a_i \), we have a series and when the upper limit of the sum is finite as in \( \displaystyle \sum_{i=1}^{n} a_i \) , we have a partitial sum .



Convergent and Divergent Series

We first discuss the convergence and the divergence divergence of series using graphs on two examples.

Example 1 - Convergent Series

Let \( a_i = \dfrac{1}{4^i} \) and define the partial sums as \( \displaystyle S_n = \sum_{i=1}^{n} \dfrac{1}{4^i} \)
The table of values of \( n \) , \( a_n \), the finite sums \( s_n \) defined by \begin{aligned} & s_1 = a_1\\ & s_2 = a_1+a_2\\ & s_3 = a_1 + a_2 + a_3\\ & s_4 = a_1 + a_2 + a_3 + a_4\\ \end{aligned} are shown below.
Note that as we add more terms to the partial sum \( s_n \), it becomes closer to \( 2 \) and that is clear from Table.1 and Graph.1 below.
table of a Convergent Series
Table.1 - Table of Values of \(n , a_n\) and the Partial Sums \( \displaystyle \sum_{i=1}^{n} \dfrac{1}{4^i} \)

The plot of \( s_n \) for \( n=1, 2, 3... \) against \( n \) is shown below and we notice that as \( n \) increases, \( s_n \) tends to a constant value equal to \( 2 \).
Graph of a Convergent Series
Graph.1 - Graph of a Partial Sums \( \displaystyle \sum_{i=1}^{n} \dfrac{1}{4^i} \)

It can be said that as \( n \) increases, the partial sum \( \displaystyle \sum_{i=1}^{n} \dfrac{1}{4^i} \) tends to \( 2 \) or has a limit equal to \( 2 \) and we therefore might think of the series \( \displaystyle \sum_{i=1}^{\infty} \dfrac{1}{4^i} \) as
convergent to \( 2 \).


Example 2 - Divergent Series

Let \( a_i = (-2)^i \) and define the partial sums as \( S_n = \displaystyle \sum_{i=1}^{n} (-2)^i \)
Below is shown the table of values of \( n \) , \( a_n \), the finite sums \( s_n \).
As we add more terms to the partial sum \( s_n \), neither the values in Table.2 nor Graph.2 show convergence to any value.
It can be said that as \( n \) increases the partial sum \( \displaystyle \sum_{i=1}^{n} \dfrac{1}{4^i} \) has no limit and we therefore might think of the series \( \displaystyle \sum_{i=1}^{\infty} \dfrac{1}{4^i} \) as
divergent .
table of a Divergent Series
Table.2 - Table of Values of Partial Sums \( \displaystyle \sum_{i=1}^{n} (-2)^i \)


Graph of a Divergent Series
Graph.2 - Graph of Partial Sums \( \displaystyle \sum (-2)^i \)



Formal Definition of Convergent and Divergent Series

Given a series \[ \sum_{i=1}^{\infty} a_i = a_1 + a_2 + a_3 + ... \] Let \( S_n \) be the partial sum \[ S_n = \sum_{i=1}^{n} a_i = a_1 + a_2 + ... + a_n \] If \( \lim_{n\to\infty} S_n \) exists and \[ \lim_{n\to\infty} S_n = s \] where \( s \) is a real number; we say that the series \( \displaystyle \sum_{i=1}^{\infty} a_i \) is convergent and write \[ \sum_{i=1}^{\infty} a_i = a_1 + a_2 + a_3 +..... = s \] If \( \lim_{n\to\infty} S_n \) does not exist or is not a real number, the series \( \sum_{i=1}^{\infty} a_i \) is divergent.

We can also write that \[ \sum_{i=1}^{\infty} a_i = \lim_{n\to\infty} \sum_{i=1}^{n} a_i \]



Geometric Series

The geometric series is defined as follows \[ \sum_{i=1}^{\infty} a \; r^{i-1} = a + a \; r + a \; r^2 +.... \] where \( r \) is called the common ratio. Let \( S_n \) be the partial sum defined by \[ S_n = a + a \; r + a \; r^2 + ... + a \; r^{n-1} \quad (I) \] Multiply both sides of the above by the common ratio \( r \) to obtain \[ r \; S_n = a \; r + a \; r^2 + a \; r^3 + ... + a \; r^n \quad (II) \] Subtract side by side (I) and (II) above to obtain \[ s_n - r \; S_n = (a + a \; r + a \; r^2 + ... + a \; r^{n-1}) - (a \; r + a \; r^2 + a \; r^3 + ... + a \; r^n) \] Simplify the right side and factor both sides \[ s_n(1 - r)= a(1 - r^n) \] Solve the above for \( S_n \) \[ S_n = \dfrac{a(1 - r^n) }{1 - r} \] It is known from limits that \( \quad \lim_{n\to\infty} r^n = 0 \) if \( \quad |r| \lt 1 \)
We can therefore state the following:
A geometric series given by \[ \sum_{i=1}^{\infty} a r^{i-1} = a + a r + a r^2 + ... \] is convergent if the common ratio \( r \) is such that \( |r| \lt 1 \) and its sum is given by \[ \boxed { \sum_{i=1}^{\infty} a r^{i-1} = \dfrac{ a }{1 - r} \quad , \quad |r| \lt 1 } \qquad (III)\] If \( |r| \gt 1 \) the geometric series is divergent.



Example 3 - Geometric Series

Which of the following geometric series are convergent? Find the sum if possible.

  1. \( 4 - \dfrac{12}{5} + \dfrac{36}{ 25} - \dfrac{108}{125} ... \)

  2. \( \displaystyle \sum_{i=1}^{\infty} \dfrac{3^{2 i}}{10^{i-1}} \)

  3. \( \displaystyle \sum_{i=1}^{\infty} \dfrac{4}{5}\cdot \dfrac{1}{5^{-i} \; 2^{i+1}} \)

Solutions to Example 3

  1. The series \[ 4 - \dfrac{12}{5} + \dfrac{36}{ 25} - \dfrac{108}{125} ... \] has the common ration \( r = \dfrac{- \dfrac{12}{5}}{4} = - \dfrac{3}{5} \)
    \( |r| = \dfrac{3}{5} \) and since \( |r| \lt 1 \) the given series is convergent.
    The first term of the series is \( a = 4 \) and using the formula (III), we obtain \[ 4 - \dfrac{12}{5} + \dfrac{36}{ 25} - \dfrac{108}{125} ... = \dfrac{4}{1 + \dfrac{3}{5} } = \dfrac{5}{2} \]


  2. Rewrite the given series as \begin{aligned} & \displaystyle \sum_{i=1}^{\infty} \dfrac{3^{2 i}}{10^{i -1}} = \sum_{i=1}^{\infty} \dfrac{(3^2)^i} {10^{i-1}}\\[15pt] & \text {Rewrite the numerator as follows} \\[15pt] & = \sum_{i=1}^{\infty} \dfrac{(3^2) (3^2)^{i-1}} {10^{i-1}}\\[15pt] & \text {which may be written as} \\[15pt] & = \sum_{i=1}^{\infty} 9 \left(\dfrac{9}{10}\right)^{i-1}\\[15pt] \end{aligned} The first term \( a = 9 \) and the common ratio \( r = \dfrac{9}{10} \), therefore \( |r| \lt 1 \) and the series is convergent and is given by. \[ \displaystyle \sum_{i=1}^{\infty} \dfrac{3^{2 i}}{10^{i -1}} = \dfrac{9}{1 - \dfrac{9}{10}} = 90 \]


  3. \begin{aligned} & \text{Rewrite the series with powers of \( 2 \) and \( 5 \)} \\[15pt] & \displaystyle \sum_{i=1}^{\infty} \dfrac{4}{5}\cdot \dfrac{1}{5^{-i} \; 2^{i+1}} = \displaystyle \sum_{i=1}^{\infty} \dfrac{2^2}{5^1}\cdot \dfrac{1}{5^{-i} \; 2^{i+1}} \\[15pt] & \text{Simplify the powers with same base and write with positive powers} \\[15pt] & = \displaystyle \sum_{i=1}^{\infty} \dfrac{5^{i - 1}}{ 2^{i-1}} \\[15pt] & \text{Rewrite as } \\[15pt] & = \displaystyle \sum_{i=1}^{\infty} \left(\dfrac{5}{2}\right)^{i - 1} \\[15pt] \end{aligned} The common ratio of the given geometric series \( r = \dfrac{5}{2} \) and therefore \( |r| \gt 1 \) hence the series is divergent.



Arithmetic Series Are Divergent

An arithmetic series is given by \[ \sum_{i=1}^{\infty} (a + (i - 1) d) = a + (a + d) + (a + 2 d) + ... \] Let
\( S_n = \displaystyle \sum_{i=1}^{n} (a + (i - 1) d ) \)
Hence
\( S_n = a + (a + d) + (a + 2 d) + ... + (a + (n-1)d ) \qquad (I) \)
Rewrite \( S_n \) by reversing the order, from last term to the first term, in the above sum.
\( S_n = (a + (n-1)d ) + (a + (n-2)d ) + (a + (n-3)d ) + ... (a+d) + a ) \qquad (II) \)
Note that:
1) adding the first term in (I) and (II), we obtain \( a + (a + (n-1)d ) = 2 a + (n-1) d \)
2) Adding the second term in (I) and (II), we obtain \( a + d + (a + (n-2)d ) = 2 a + (n-1) d \)
3) 2) Adding the third term in (I) and (II), we obtain \( a + 2 d + (a + (n-3)d ) = 2 a + (n-1) d \)
Noting that there \( n \) terms on the right sides of (I) and (II) and adding (I) and (II) term by term we obtain
\( 2 S_n = n ( 2 a + (n-1) d ) \)
and
\[ S_n = \dfrac{ n ( 2 a + (n-1) d )}{2} \]
Expand the numerator and write
\[ S_n = \dfrac{n^2d}{2} + n (a -\dfrac{d}{2} ) \]
Note that \[ \lim_{n\to\infty} (\dfrac{n^2d}{2} + n (a -\dfrac{d}{2} )) = \infty \] and therefore all arithmetic series are divergent.



Theorem of Combinations of Convergent Series

If \( \displaystyle \sum_{n=1}^{\infty} a_n \) and \( \displaystyle \sum_{n=1}^{\infty} b_n \) are convergent series such that \( \displaystyle \sum_{n=1}^{\infty} a_n = L_1\) and \( \displaystyle \sum_{n=1}^{\infty} b_n = L_2\) and \( k_1 \) and \( k_2 \) are real numbers, then any linear combination of the given series of the form \[ k_1 \sum_{n=1}^{\infty} a_n + k_2 \sum_{n=1}^{\infty} b_n \] is convergent and is given by \[ k_1 \sum_{n=1}^{\infty} a_n + k_2 \sum_{n=1}^{\infty} b_n = k_1 L_1 + k_2 L_2 \]



Example 4

Evaluate, if possible, series:

  1. \( (2 - 1 + \dfrac{1}{ 2} - \dfrac{1}{4} + ...) + ( - \dfrac{1}{ 3} + \dfrac{1}{ 9} - \dfrac{1}{ 27} + ... ) \)

  2. \( - 3 \displaystyle \sum_{i=1}^{\infty} \left(\dfrac{1}{2} \right)^{i-1} - 4 \sum_{i=1}^{\infty} 0.7^{i-1} \)

  3. \( \displaystyle \sum_{i=1}^{\infty} \left(0.2 \right)^{i} + \sum_{i=1}^{\infty} \dfrac{(3^i - 7 \cdot 2^i)}{5^{i-1}} \)

Solutions to Example 4

  1. \begin{aligned} & \text{The series \( 2 - 1 + \dfrac{1}{ 2} - \dfrac{1}{4} + ... \) is geometric with first term \(A_1 = 2\) and common ratio \( R_1 = - \dfrac{1}{ 2} \) and is therefore convergent since \( |R_1| \lt 1 \).}\\[15pt] & \text{The series \( - \dfrac{1}{ 3} + \dfrac{1}{ 9} - \dfrac{1}{ 27} + ... \) is geometric with first term \(A_2 = - \dfrac{1}{ 3} \) and common ratio \( R_2 = - \dfrac{1}{ 3} \) and is convergent since \( |R_2| \lt 1 \).}\\[15pt] & \text{ Use formula (III) for each series and the theorem of the combination above to write} \\[15pt] & (2 - 1 + \dfrac{1}{ 2} - \dfrac{1}{4} + ...) + ( - \dfrac{1}{ 3} + \dfrac{1}{ 9} - \dfrac{1}{ 27} + ... ) = \dfrac{A_1}{1 - R_1 } + \dfrac{A_2}{1 - R_2} \\[15pt]
    & \text{ Substitute \( A_1, A_2, R_1\) and \( R_2 \) by their values}\\[15pt]
    & = \dfrac{2}{1 + \dfrac{1}{ 2} } + \dfrac{- \dfrac{1}{ 3}}{1 + \dfrac{1}{ 3} } \\[15pt]
    & Simplify \\[15pt]

    & = \dfrac{13}{12} \end{aligned}



  2. \begin{aligned} & \text{The series \( \displaystyle \sum_{i=1}^{\infty} \left(\dfrac{1}{2} \right)^{i-1} \) and \( \sum_{i=1}^{\infty} 0.7^{i-1} \) have first terms \(A_1 = 1 \) and \(A_2 = 1 \) and common ratios \(R_1 = \dfrac{1}{2} \) and \(R_2 = 0.7 \) }\\[15pt] & \text{Both common ratios have absolute values less than \( 1 \) and therefore the two series are convergent} \\[15pt] & \text{ Use formula (III) for each series and the theorem of the combination above to obtain} \\[15pt]
    & - 3 \displaystyle \sum_{i=1}^{\infty} \left(\dfrac{1}{2} \right)^{i-1} - 4 \sum_{i=1}^{\infty} 0.7^{i-1} = - 3 \dfrac{A_1}{1-R_1} - 4 \dfrac{A_2}{1 - R_2} \\[15pt] & \text{ Substitute} \\[15pt] & = - 3 \dfrac{1}{1-\dfrac{1}{2}} - 4 \dfrac{1}{1 - 0.7} \\[15pt] &\text{Simplify} \\[15pt] & = -\dfrac{58}{3} \end{aligned}



  3. \begin{aligned} & \text{Given} \\[15pt] & \displaystyle \sum_{i=1}^{\infty} \left(0.2 \right)^i + \sum_{i=1}^{\infty} \dfrac{(3^i - 7 \cdot 2^i)}{5^{i-1}} \\[15pt] & \text{Rewite the given expression as a combination of three geometric series as follows}\\[15pt] & = 0.2 \sum_{i=1}^{\infty} \left(0.2 \right)^{i-1} + 3 \sum_{i=1}^{\infty} \left(\dfrac{3}{5}\right)^{i-1} - 14 \sum_{i=1}^{\infty} \left(\dfrac{2}{5} \right)^{i-1} \\[15pt] & \text{Identify the three geometric series \( \sum_{i=1}^{\infty} \left(0.2 \right)^{i-1} \), \( \sum_{i=1}^{\infty} \left(\dfrac{3}{5}\right)^{i-1} \) and \( \sum_{i=1}^{\infty} \left(\dfrac{2}{5} \right)^{i-1} \) as having the }\\[15pt] & \text{ first term equal to \( 1 \) and the common ratios equal to \( 0.2 , \; \dfrac{3}{5} , \; \dfrac{2}{5} \) respectively } \\[15pt] & \text{The three geometric series have ratios whose absolute values are less than \( 1 \) and are therefore convergent } \\[15pt] & \text{Use formula (III) for each series and the theorem of the combination above, to obtain } \\[15pt] & = 0.2 \dfrac{1}{1-0.2} + 3 \dfrac{1}{1 - \dfrac{3}{5}} - 14 \dfrac{1}{1-\dfrac{2}{5}} \\[15pt] & \text{Simplify} \\[15pt] & = -\frac{187}{12} \end{aligned}



    More References and Links

    1. University Calculus - Early Transcendental - Joel Hass, Maurice D. Weir, George B. Thomas, Jr., Christopher Heil - ISBN-13 : 978-0134995540
    2. Calculus - Early Transcendental - James Stewart - ISBN-13: 978-0-495-01166-8
    3. Sequences and Summation
    4. Introduction to Limits

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