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.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.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.2 - Table of Values of Partial Sums \( \displaystyle \sum_{i=1}^{n} (-2)^i \)
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.
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
\[ 4 - \dfrac{12}{5} + \dfrac{36}{ 25} - \dfrac{108}{125} ... = \dfrac{4}{1 + \dfrac{3}{5} } = \dfrac{5}{2} \]
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.
\[ \displaystyle \sum_{i=1}^{\infty} \dfrac{3^{2 i}}{10^{i -1}} = \dfrac{9}{1 - \dfrac{9}{10}} = 90 \]
\[
\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
\[ k_1 \sum_{n=1}^{\infty} a_n + k_2 \sum_{n=1}^{\infty} b_n = k_1 L_1 + k_2 L_2 \]
\[
\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} \]
\[
\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} \]
\[
\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
University Calculus - Early Transcendental - Joel Hass, Maurice D. Weir, George B. Thomas, Jr., Christopher Heil - ISBN-13 : 978-0134995540
Calculus - Early Transcendental - James Stewart - ISBN-13: 978-0-495-01166-8
