Infinite Series

An infinite series is a sum with infinitely many terms.

If a sequence has terms \(u_1, u_2, u_3, \dots\), then the \(n^{\text{th}}\) partial sum is

\[ S_n = u_1 + u_2 + \dots + u_n. \]

If \(S_n\) tends to a limit \(S\) as \(n \to \infty\), then \(S\) is called the sum to infinity of the series:

\[ \sum_{n=1}^{\infty} u_n = S \quad\text{if}\quad \lim_{n\to\infty} S_n = S. \]

Arithmetic Series

For an arithmetic sequence with first term \(a\) and common difference \(d\):

\[ u_n = a + (n-1)d \] \[ S_n = \frac{n}{2}\bigl(2a + (n-1)d\bigr). \]

As \(n \to \infty\), unless \(d = 0\), the partial sums grow without bound.

Conclusion: the sum to infinity of a non‑constant arithmetic series is undefined.

Geometric Series

For a geometric sequence with first term \(a\) and common ratio \(r\):

\[ u_n = ar^{n-1} \]

The sum of the first \(n\) terms is:

\[ S_n = a\,\frac{1 - r^n}{1 - r}, \quad r \ne 1. \]

Behaviour of \(r^n\) as \(n \to \infty\):

if \(|r| > 1\), then \(r^n \to \infty\) or diverges in magnitude;
if \(|r| < 1\), then \(r^n \to 0\).

The sum to infinity is:

Undefined when \(|r| \ge 1\).

Defined when \(|r| < 1\), and then

\[ \sum_{n=1}^{\infty} ar^{n-1} = \frac{a}{1-r}, \quad |r| < 1. \]

Example

Find the sum to infinity of the series

\(96 + 48 + 24 + \dots\), if it exists.

Here \(a = 96\), \(r = \dfrac{48}{96} = \dfrac{1}{2}\), and \(|r| < 1\).

\[ S_{\infty} = \frac{96}{1 - \frac{1}{2}} = \frac{96}{\frac{1}{2}} = 192. \]

Example

Express the recurring decimal \(0.242424\ldots\) as a fraction.

Write \[ 0.242424\ldots = 0.24 + 0.0024 + 0.000024 + \dots \] This is geometric with \[ a = 0.24,\quad r = 0.01. \]

\[ S_{\infty} = \frac{0.24}{1 - 0.01} = \frac{0.24}{0.99} = \frac{24}{99} = \frac{8}{33}. \]

Example

Given that 12 and 6 are two adjacent terms of an infinite geometric series with sum to infinity \(192\):

(a) Find the first term. (b) Find the partial sum \(S_6\).

Adjacent terms 12 and 6 give \[ r = \frac{6}{12} = \frac{1}{2}. \]

Let the first term be \(a\). Then

\[ S_{\infty} = \frac{a}{1-r} = 192 \quad\Rightarrow\quad \frac{a}{1 - \frac{1}{2}} = 192 \quad\Rightarrow\quad a = 96. \]

The terms are \(96, 48, 24, 12, 6, 3, \dots\)

\[ S_6 = 96\,\frac{1 - \left(\frac{1}{2}\right)^6}{1 - \frac{1}{2}} = 96 \cdot 2 \left(1 - \frac{1}{64}\right) = 192 \cdot \frac{63}{64} = \frac{12096}{64} = 189. \]

Sigma Notation – Rules

We use sigma notation to write sums compactly:

\[ \sum_{k=1}^{n} u_k = u_1 + u_2 + \dots + u_n. \]

Constant multiplier:

\[ \sum_{k=1}^{n} c\,u_k = c \sum_{k=1}^{n} u_k. \]

Adding series:

\[ \sum_{k=1}^{n} (u_k + v_k) = \sum_{k=1}^{n} u_k + \sum_{k=1}^{n} v_k. \]

Sum of First \(n\) Natural Numbers

The first \(n\) natural numbers are:

\[ 1 + 2 + 3 + \dots + n. \]

\[ \sum_{r=1}^{n} r = 1 + 2 + 3 + 4 + 5 + \ldots + n \] \[ \text{written backwards} \] \[ \sum_{r=1}^{n} r = n + (n-1) + (n-2) + \ldots + 3 + 2 + 1 \] \[ \text{added} \] \[ 2\sum_{r=1}^{n} r = (1+n) + (2+n-1) + (3+n-2) + \ldots + (n+1) \] \[ = (n+1) + (n+1) + (n+1) + \ldots + (n+1) \] \[ \therefore\quad 2\sum_{r=1}^{n} r = n(n+1) \] \[ \Rightarrow\quad \sum_{r=1}^{n} r = \frac{n(n+1)}{2} \]

\[ \sum_{k=1}^{n} k = \frac{n(n+1)}{2}. \]

\[ \sum_{r=1}^{n} (ar + b) = (a + b) + (2a + b) + (3a + b) + \ldots + (na + b) \] \[ = a + 2a + 3a + \ldots + na \;+\; b + b + b + \ldots + b \] \[ = a(1 + 2 + 3 + \ldots + n) \;+\; b(1 + 1 + 1 + \ldots + 1) \] \[ = a\sum_{r=1}^{n} r \;+\; b\sum_{r=1}^{n} 1 \] \[ = a \cdot \frac{n(n+1)}{2} \;+\; b \cdot n \] \[ = \frac{1}{2} a n (n+1) + bn \]

Example

Find the value of

\[ \sum_{r=1}^{n} (5r + 3) \] \[ = 5 + 3 \;+\; 10 + 3 \;+\; 15 + 3 \;+\; \ldots \;+\; 5n + 3 \] \[ = 5(1 + 2 + 3 + \ldots + n) \;+\; 3(1 + 1 + 1 + \ldots + 1) \] \[ = 5\sum_{r=1}^{n} r \;+\; 3\sum_{r=1}^{n} 1 \] \[ = 5 \cdot \frac{n(n+1)}{2} \;+\; 3n \] \[ = \frac{5}{2}n(n+1) + 3n \] \[ = \frac{5}{2}n^2 + \frac{11}{2}n \] \[ \boxed{\sum_{r=1}^{n} (5r + 3) = \frac12\,n(5n + 11)} \]

Useful results:

\[ \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}, \] \[ \sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2. \]

Example

Find the value of

\[ \sum_{k=1}^{10} (3k^2 + 2k). \]

Using the rules:

\[ \sum_{k=1}^{10} (3k^2 + 2k) = 3\sum_{k=1}^{10} k^2 + 2\sum_{k=1}^{10} k. \] \[ 3\cdot \frac{10\cdot 11\cdot 21}{6} + 2\cdot \frac{10\cdot 11}{2} = 3\cdot 385 + 2\cdot 55 = 1155 + 110 = 1265. \]

Common Series – Sigma Notation

Some common finite sums:

\[ \text{Even numbers} \] \[ \sum_{r=1}^{n} 2r = 2 + 4 + 6 + 8 + 10 + \ldots + 2n \]

\[ \text{Odd numbers} \] \[ \sum_{r=1}^{n} (2r - 1) = 1 + 3 + 5 + 7 + \ldots + (2n - 1) \] \[ \text{or} \] \[ \sum_{r=1}^{n} (2r + 1) = 3 + 5 + 7 + \ldots + (2n + 1) \]

\[ \sum_{r=1}^{n} r = 1 + 2 + 3 + 4 + 5 + \ldots + n \]

\[ \sum_{k=1}^{n} 1 = n, \] \[ \sum_{k=1}^{n} k = \frac{n(n+1)}{2}, \] \[ \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}, \] \[ \sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2. \]

\[ \sum_{r=1}^{n} x^{r} = 1 + x + x^{2} + x^{3} + x^{4} + x^{5} + \ldots + x^{n} \]

\[ \sum_{r=1}^{n} (-1)^{r} = (-1)^{1} + (-1)^{2} + (-1)^{3} + (-1)^{4} + (-1)^{5} + \ldots + (-1)^{n} \] \[ = -1 + 1 - 1 + 1 - 1 + \ldots + (-1)^{n} \] \[ \sum_{r=1}^{n} (-1)^r = \begin{cases} 0, & \text{if } n \text{ is even},\\[6pt] -1, & \text{if } n \text{ is odd}. \end{cases} \] \[ \sum_{r=1}^{n} (-1)^r = -\frac{1 - (-1)^n}{2}. \]

\[ \sum_{r=1}^{n} (-1)^{r+1} = (-1)^{1+1} + (-1)^{2+1} + (-1)^{3+1} + \ldots + (-1)^{n+1} \] \[ = 1 - 1 + 1 - 1 + 1 - \ldots + (-1)^{n+1} \] \[ \text{Closed form:}\quad \sum_{r=1}^{n} (-1)^{r+1} = \begin{cases} 0, & \text{if } n \text{ is even},\\[4pt] 1, & \text{if } n \text{ is odd}. \end{cases} \] \[ \text{Equivalently:}\quad \sum_{r=1}^{n} (-1)^{r+1} = \dfrac{1 - (-1)^n}{2}. \]

\[ \frac{1}{1 - x} = \sum_{r=0}^{\infty} x^{r} = 1 + x + x^{2} + x^{3} + x^{4} + x^{5} + \ldots + x^{n-1}, \quad -1 \lt x \lt 1 \]

\[ \sum_{k=0}^{n-1} ar^k = a\,\frac{1 - r^n}{1 - r},\quad r \ne 1. \]

Power Series

A power series in \(x\) has the form:

\[ \sum_{n=0}^{\infty} a_n x^n = a_0 + a_1 x + a_2 x^2 + \dots \]

where \(a_0, a_1, a_2, \dots\) are constants.

The series always converges at \(x = 0\). It may converge for other values of \(x\).

A necessary condition for convergence is that the terms tend to zero: \[ \lim_{n\to\infty} a_n x^n = 0. \]

D’Alembert’s Ratio Test

For a series of positive terms \(\sum u_n\), consider

\[ L = \lim_{n\to\infty} \frac{u_{n+1}}{u_n}. \]

if \(L < 1\), the series converges;
if \(L > 1\), the series diverges;
if \(L = 1\), the test is inconclusive.

Example

Use d’Alembert’s ratio test to test convergence of

\[ \sum_{n=1}^{\infty} \frac{n!}{2^n}. \]

Here \(u_n = \dfrac{n!}{2^n}\). Then

\[ \frac{u_{n+1}}{u_n} = \frac{(n+1)!/2^{n+1}}{n!/2^n} = \frac{n+1}{2}. \]

As \(n \to \infty\), \(\dfrac{n+1}{2} \to \infty > 1\), so the series diverges.

Absolute Convergence

A series \(\sum u_n\) is said to be absolutely convergent if

\[ \sum |u_n| \]

converges.

If a series is absolutely convergent, then it is convergent.

\[ \text{If}\quad \sum \lvert u_n \rvert \ \text{is divergent and}\quad \sum u_n \ \text{converges,} \] \[ \text{then}\quad \sum u_n \ \text{is conditionally convergent}. \]

\[ \lim_{n \to \infty} \left| \frac{a_{n+1} x}{a_n} \right| > 1 \quad \Rightarrow \quad \text{series diverges or converges conditionally.} \]

\[ \lim_{n \to \infty} \left| \frac{a_{n+1} x}{a_n} \right| = 1 \quad \Rightarrow \quad \text{indeterminate}. \]

Example

Example. Find the sum to infinity of the series

\[ 4 + 7x + 10x^{2} + 13x^{3} + \ldots \]

and the region of valid values of \(x\).

\[ S_n = 4 + 7x + 10x^{2} + 13x^{3} + \ldots + (3n+1)x^{\,n-1} \]

Multiply by \(x\):

\[ xS_n = 4x + 7x^{2} + 10x^{3} + 13x^{4} + \ldots + (3n+1)x^{\,n} \]

Subtract:

\[ S_n(1 - x) = 4 + 3x + 3x^{2} + 3x^{3} + \ldots + 3x^{\,n-1} - (3n+1)x^{\,n} \] \[ = 4 + 3\left(x + x^{2} + x^{3} + \ldots + x^{\,n-1}\right) - (3n+1)x^{\,n} \] \[ = 4 + 3\left(\frac{x(1 - x^{\,n-1})}{1 - x}\right) - (3n+1)x^{\,n} \]

\[ \text{If } -1 \lt x \lt 1,\ \text{then as } n \to \infty,\quad x^{n} \to 0,\quad x^{\,n-1} \to 0. \] \[ \Rightarrow\ S_{\infty}(1 - x) = 4 + 3\left(\frac{x}{1 - x}\right) \] \[ \Rightarrow\ S_{\infty}(1 - x) = \frac{4(1 - x) + 3x}{1 - x} \] \[ \Rightarrow\ S_{\infty}(1 - x) = \frac{4 - x}{1 - x} \] \[ \Rightarrow\ S_{\infty} = \frac{4 - x}{(1 - x)^{2}} \qquad\text{for } -1 \lt x \lt 1. \]

Fibonacci Series

The Fibonacci sequence is defined by:

\[ u_1 = 1,\quad u_2 = 1,\quad u_{n+2} = u_{n+1} + u_n. \]

So the sequence begins:

\[ 1,\; 1,\; 2,\; 3,\; 5,\; 8,\; 13,\; \dots \]

\[ \text{If } F_n \text{ is the } n^{\text{th}} \text{ Fibonacci number,} \] \[ \lim_{n \to \infty} \frac{F_{n+1}}{F_n} = \frac{1 + \sqrt{5}}{2} \] \[ \lim_{n \to \infty} \left| \frac{F_{n+1} x^{\,n+1}}{F_n x^{\,n}} \right| \lt 1 \quad\Rightarrow\quad \frac{1 + \sqrt{5}}{2}\, |x| \lt 1 \] \[ \text{So the series converges when } |x| \lt \frac{2}{1 + \sqrt{5}}. \]

Why does the ratio of successive Fibonacci numbers approach the golden ratio?

The Fibonacci sequence satisfies the recurrence

\[ F_{n+1} = F_n + F_{n-1}. \]

Divide both sides by \(F_n\):

\[ \frac{F_{n+1}}{F_n} = 1 + \frac{F_{n-1}}{F_n} = 1 + \frac{1}{F_n/F_{n-1}}. \]

Let

\[ L = \lim_{n\to\infty} \frac{F_{n+1}}{F_n}. \]

Then also

\[ \lim_{n\to\infty} \frac{F_n}{F_{n-1}} = L. \]

So in the limit the recurrence becomes

\[ L = 1 + \frac{1}{L}. \]

Multiply through:

\[ L^2 = L + 1. \]

Solve the quadratic:

\[ L = \frac{1 \pm \sqrt{5}}{2}. \]

Since Fibonacci ratios are positive, we take the positive root:

\[ \boxed{ \displaystyle \lim_{n\to\infty} \frac{F_{n+1}}{F_n} = \frac{1 + \sqrt{5}}{2} } \]

This number is the golden ratio.

Centre of Convergence

A power series can be written in the form:

\[ \sum_{n=0}^{\infty} a_n (x - c)^n \] \[ = \lim_{n \to \infty} \bigl( a_0 + a_1(x - c) + a_2(x - c)^2 + a_3(x - c)^3 + \ldots + a_n (x - c)^n \bigr). \]

where \(c\) is the centre of convergence.

This is the middle of the interval of convergence, the interval for which the limit exists.
The series converges for \(x\) in an interval centred at \(c\), of radius \(R\), called the radius of convergence.

If \(R = \infty\), the series converges for all real \(x\).

Otherwise, the series converges for \( |x - c| < R \)

and diverges for \( |x - c| > R. \)

\[ f(x) = a_0 + a_1(x - c) + a_2(x - c)^2 + a_3(x - c)^3 + \ldots \]

To find a₀, set x = c , since the remaining terms will become zero.

differentiate

\[ f'(x) = a_1 + 2a_2(x - c) + 3a_3(x - c)^2 + 4a_4(x - c)^3 + \ldots \]

To find a₁, evaluate f’(c)

go again

\[ f''(x) = 2a_2 + 3\cdot 2\, a_3 (x - c) + 4\cdot 3\, a_4 (x - c)^2 + \ldots \] \[ f''(c) = 2a_2 \] \[ \Rightarrow\quad a_2 = \frac{f''(c)}{2} \] \[ \Rightarrow\quad a_2 = \frac{f''(c)}{2!} \]

Continuing gives the following

\[ f'''(x) = 3\cdot 2\, a_3 + 4\cdot 3\cdot 2\, a_4 (x - c) + 5\cdot 4\cdot 3\, a_5 (x - c)^2 + \ldots \] \[ f'''(c) = 3\cdot 2\, a_3 \] \[ \Rightarrow\quad a_3 = \frac{f'''(c)}{3\cdot 2} \] \[ \Rightarrow\quad a_3 = \frac{f'''(c)}{3!} \]

and

\[ f^{iv}(x) = 4\cdot 3\cdot 2\, a_4 + 5\cdot 4\cdot 3\cdot 2\, a_5 (x - c) + \ldots \] \[ f^{iv}(c) = 4\cdot 3\cdot 2\, a_4 \] \[ \Rightarrow\quad a_4 = \frac{f^{iv}(c)}{4\cdot 3\cdot 2} \] \[ \Rightarrow\quad a_4 = \frac{f^{iv}(c)}{4!} \]

so that

\[ a_n = \frac{f^{(n)}(c)}{n!} \]

Substituting back into the original series

\[ f(x) = a_0 + a_1(x - c) + a_2(x - c)^2 + a_3(x - c)^3 + \ldots \] \[ = f(c) + f'(c)(x - c) + \frac{f''(c)(x - c)^2}{2} + \frac{f'''(c)(x - c)^3}{3\cdot 2} + \ldots \] \[ = f(c) + f'(c)(x - c) + \frac{f''(c)(x - c)^2}{2!} + \frac{f'''(c)(x - c)^3}{3!} + \ldots \]

\[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!}\,(x - c)^n \] \[ = f(c) + f'(c)(x - c) + \frac{f''(c)(x - c)^2}{2!} + \frac{f'''(c)(x - c)^3}{3!} + \ldots \]

In the interval \((\, -R + c,\; R + c \,)\)

\[ f'(x) = \sum_{n=1}^{\infty} n a_n (x - c)^{n-1} \quad\Rightarrow\quad f'(c) = a_1, \] \[ f''(x) = \sum_{n=2}^{\infty} n(n-1)a_n (x - c)^{n-2} \quad\Rightarrow\quad f''(c) = 2! a_2, \]

and in general

\[ f^{(n)}(c) = n! a_n \quad\Rightarrow\quad a_n = \frac{f^{(n)}(c)}{n!}. \]

Taylor’s Series

The Taylor series of a function \(f(x)\) about \(x = c\) is

\[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!}(x - c)^n. \]

In particular, about \(x = 0\) (Maclaurin series):

\[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \dots \]

For worked examples, see the Taylor / Maclaurin examples .