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Natural Sciences Notes Mathematics for Natural Sciences Notes

Sequences And Series Notes

Updated Sequences And Series Notes

Mathematics for Natural Sciences Notes

Mathematics for Natural Sciences

Approximately 176 pages

Notes for every topic covered in Mathematics for Natural Sciences, Part IA. The notes emphasise the most important aspects of each topic - specifically the material which is commonly addressed in examination questions.

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Notes for Sequences and Series

General form of infinite series:

An infinite series is a summation of an infinite number of terms. It can be expressed in terms of partial sums (Sn) of the form:

$$S_{n} = \sum_{k = 0}^{n}u_{k}$$

Where Sn is a finite summation to the nth term of the sequence.

Convergence:

Infinite series are either convergent or divergent:

Convergent series: Divergent series:
Partial sums have a finite limit SnS as n Partial sums diverge Sn as n

Definition of convergence:
Formally, a series is convergent (limnSnS) if it satisfies the following condition:

$$\boxed{\underset{\mathbf{n \rightarrow \infty}}{\mathbf{\lim}}\mathbf{S}_{\mathbf{n}}\mathbf{= S \Longleftrightarrow}\mathbf{for\ any\ }\mathbf{\epsilon,\exists N}\mathbf{\ such\ that\ }\left| \mathbf{S -}\mathbf{S}_{\mathbf{n}} \right|\mathbf{< \epsilon\ }\mathbf{for\ all}\mathbf{\ n > N}}$$

Properties of convergence:

  1. Convergence is a statement of behaviour as n, thus altering a finite number of terms in an infinite series does not change its convergence properties

  2. The condition that uk0 as k is necessary for convergence, however is not sufficient to guarantee convergence

  3. The sum of convergent series is always convergent: If two convergent series are defined such that $\sum_{k = 0}^{\infty}u_{k} = A$ and $\sum_{k = 0}^{\infty}v_{k} = B$, then $\sum_{k = 0}^{\infty}\left( u_{k} + v_{k} \right) = A + B$

  4. The sum of divergent series may also be convergent: If a convergent series is defined such that $\sum_{k = 0}^{\infty}\left( u_{k} + v_{k} \right) = C$, then the two series $\sum_{k = 0}^{\infty}u_{k}$ and $\sum_{k = 0}^{\infty}v_{k}$ are not necessarily convergent.

Types of convergence:

Convergent series may be absolutely or conditionally convergent:

Absolutely convergent series: Conditionally convergent series:

The series defined by $S_{n} = \sum_{k = 0}^{n}u_{k}$ is absolutely convergent if $A_{n} = \sum_{k = 0}^{n}\left| u_{k} \right|$ is convergent:

$$\boxed{\underset{\mathbf{n \rightarrow \infty}}{\mathbf{\lim}}{\sum_{\mathbf{k = 0}}^{\mathbf{n}}\left| \mathbf{u}_{\mathbf{k}} \right|}\mathbf{= A}}$$

A series which fulfils this condition converges even when the absolute value of its terms are taken – making it most likely to diverge. Thus absolute convergence indicates that the series is convergent in any circumstance.

The series is convergent, but not absolutely:

$$\lim_{n \rightarrow \infty}{\sum_{k = 0}^{n}u_{k}} = S,\ \ \lim_{n \rightarrow \infty}{\sum_{k = 0}^{n}\left| u_{k} \right|} = \pm \infty$$

The series can be arranged in such a way that its convergence properties are altered, by grouping terms in a specific order, so that the number of positive terms for each negative term is increased, or vice versa.

Rearranging a conditionally convergent series:

$1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \ldots$ is a conditionally-convergent series. When its terms are grouped as $1 - \left( \frac{1}{2} - \frac{1}{3} \right) - \left( \frac{1}{4} - \frac{1}{5} \right) - \ldots$, then the series converges to ln2. When the terms are grouped $\left( 1 + \frac{1}{3} + \frac{1}{5} \right) - \frac{1}{2} + \left( \frac{1}{7} + \frac{1}{9} + \frac{1}{11} + \frac{1}{13} + \frac{1}{15} \right) - \frac{1}{4} + \left( \frac{1}{17} + \ldots + \frac{1}{25} \right) - \frac{1}{6}$, however, the number of positive terms for each negative term is increased, and hence the limit of convergence changes to $\frac{3}{2}$.

Common infinite series:

Geometric series: Riemann-Zeta series:

Defined by partial sums:

$$\mathbf{S}_{\mathbf{n}}\mathbf{=}\sum_{\mathbf{k = 0}}^{\mathbf{n}}\mathbf{r}^{\mathbf{k}}$$

SnSnr=1rn+1

Sn(1r)=1rn+1

$$\mathbf{\therefore}\mathbf{S}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{1 -}\mathbf{r}^{\mathbf{n + 1}}}{\mathbf{1 - r}}$$

Defined by partial sums:

$$\mathbf{S}_{\mathbf{n}}\mathbf{=}\sum_{\mathbf{k = 1}}^{\mathbf{n}}\mathbf{k}^{\mathbf{-}\mathbf{p}}$$

The special case of p=1 is called the Harmonic series:

$$\mathbf{S}_{\mathbf{n}}\mathbf{=}\sum_{\mathbf{k = 1}}^{\mathbf{n}}\frac{\mathbf{1}}{\mathbf{k}}$$

Convergence properties: Convergent |r|<1, divergent |r|1:

If |r|1, then uk does not tend to zero as k, thus the series is necessarily divergent.

Convergence properties: Convergent p>1, divergent p1:

If n=1 (harmonic series), then grouping terms:

$$S_{\infty} = 1 + \frac{1}{2} + \left( \frac{1}{3} + \frac{1}{4} \right) + \left( \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} \right) + \ldots$$

Each group g contains q terms of the form:

$$g = \left( \frac{1}{q + 1} + \frac{1}{q + 2} + \frac{1}{q + 3} + \ldots + \frac{1}{q + q} \right)$$

Where each term before $\frac{1}{q + q}$ is larger than $\frac{1}{2q}$:

$$\therefore g \geq \left( \frac{1}{2q} + \frac{1}{2q} + \frac{1}{2q} + \ldots + \frac{1}{2q} \right)$$

$$\therefore g \geq q\left( \frac{1}{2q} \right) \geq \frac{1}{2}$$

Therefore by comparison, the harmonic series is divergent:

$$\sum_{k = 1}^{\infty}\frac{1}{k} > \sum_{k = 1}^{\infty}\frac{1}{2},\ \ \sum_{k = 1}^{\infty}\frac{1}{2} = \infty\therefore S_{\infty} \rightarrow \infty$$
If |r|<1, then limn|r|n+1=0 and Sn converges absolutely, such that $S_{\infty} = \frac{1}{1 - r}$.

Regrouping the terms in the series:

$$S_{\infty} = \sum_{k = 1}^{\infty}k^{- p} = 1 + \left( \frac{1}{2^{p}} + \frac{1}{3^{p}} \right) + \left( \frac{1}{4^{p}} + \ldots + \frac{1}{7^{p}} \right) + \ldots$$

$$\therefore\sum_{k = 1}^{\infty}k^{- p} < 1 + 2\left( \frac{1}{2^{p}} \right) + 4\left( \frac{1}{4^{p}} \right) + 8\left( \frac{1}{8^{p}} \right) + \ldots < 1 + \frac{1}{2^{p - 1}} + \frac{1}{4^{p - 1}} + \frac{1}{8^{p - 1}}\ $$

Therefore by comparison, the series is convergent:

$$\therefore\sum_{k = 1}^{\infty}k^{- p} < \sum_{k = 1}^{\infty}\left( \frac{1}{2^{p - 1}} \right)^{k} < \frac{1}{1 - \frac{1}{2^{p - 1}}}\therefore S_{\infty} \leq \frac{1}{1 - \frac{1}{2^{p - 1}}}$$

Convergence tests for positive series:

The...

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