# 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.

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 $$2$$.

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 .

## 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}