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4.1 Finite and Infinite Series

4.1.1 Finite Series of Natural Numbers (Gauß)

\begin{eqnarray} {\mathop{∑ }}_{k=1}^{n}k = 1 + 2 + 3 + ... + n = {n(n + 1)\over 2} .& & %&(4.1) \\ \end{eqnarray}

First Proof (C. F. Gauß) for n = 100:

\begin{eqnarray} {\mathop{∑ }}_{k=1}^{100}k = (1 + 100) + (2 + 99) + (3 + 98) + ... + (50 + 51) = 50 ⋅ 101 = 5050.& & %&(4.2) \\ \end{eqnarray}

General proof by induction:

1. Induction Start (n = 1):

\begin{eqnarray} {\mathop{∑ }}_{k=1}^{1}k = 1 = {1(1 + 1)\over 2} ⇝ OK.& & %&(4.3) \\ \end{eqnarray}

2. Induction Step (n → n + 1): Assume Eq. (4.1) is true for n, then show that it is also true for n + 1:

\begin{eqnarray} {\mathop{∑ }}_{k=1}^{n+1}k ={ \mathop{∑ }}_{k=1}^{n}k + (n + 1) = {n(n + 1)\over 2} + (n + 1) = {(n + 1)(n + 2)\over 2} ⇝ OK.& & %&(4.4) \\ \end{eqnarray}

This can now be used to prove it is true for n + 2, etc.

4.1.2 Finite Geometric Progression

\begin{eqnarray} {\mathop{∑ }}_{k=0}^{n}{x}^{k} = {1 − {x}^{n+1}\over 1 − x} ,\quad x\mathrel{≠}0.& & %&(4.5) \\ \end{eqnarray}

Proof: Write

\begin{eqnarray}{ S}_{n} ={ \mathop{∑ }}_{k=0}^{n}{x}^{k}& =& 1 + x + {x}^{2} + ... + {x}^{n} %& \\ & =& 1 + x(1 + x + {x}^{2} + ... + {x}^{n−1} + {x}^{n}) − {x}^{n+1}%& \\ & =& 1 + x{S}_{n} − {x}^{n+1} ⇝ %& \\ {S}_{n}& =& {1 − {x}^{n+1}\over 1 − x} . %&(4.6) \\ \end{eqnarray}

Alternative proof by induction: home exercise.

4.1.3 Binomial

\begin{eqnarray}{ (x + y)}^{n} ={ \mathop{∑ }}_{k=0}^{n}\left (\array{ n\cr k } \right ){x}^{n−k}{y}^{k}.& & %&(4.7)\\ \end{eqnarray}

Here, we define the binomial coefficient

\begin{eqnarray} \left (\array{ n\cr k } \right ) = {n!\over k!(n − k)!} = {n(n − 1) ⋅ ...(n − k + 1)\over 1 ⋅ 2 ⋅ ... ⋅ k} .& & %&(4.8)\\ \end{eqnarray}

The proof of Eq.(4.7) goes again via induction n → n + 1. Not shown here. Examples:

\begin{eqnarray}{ (x + y)}^{2}& =& {x}^{2} + 2xy + {y}^{3} %& \\ {(x + y)}^{3}& =& {x}^{3} + 3{x}^{2}y + 3x{y}^{2} + {y}^{3}.%&(4.9) \\ \end{eqnarray}

4.1.4 Infinite Series

Definition

A series

\begin{eqnarray} S :={ \mathop{∑ }}_{k=0}^{∞}{a}_{ k} = {a}_{0} + {a}_{1} + {a}_{2} + ... ={\mathop{ lim}}_{n→∞}{S}_{n},\quad {S}_{n} :={ \mathop{∑ }}_{k=0}^{n}{a}_{ k}& & %&(4.10) \\ \end{eqnarray}

is called infinite series. It is the limit of the sequence of finite series {S}_{n} when the upper limit n tends toward infinity. (The objects {S}_{n} are also called “partial sums”.) In contrast to the finite series {S}_{n}, the infinite series S can diverge. S is said to be convergent is {S}_{n} approaches a finite limit as n →∞.

Example 4.1: 

Use a constant {a}_{k}

\begin{eqnarray}{ a}_{k} = 1 ⇝{\mathop{∑ }}_{k=0}^{∞}{a}_{ k} = 1 + 1 + 1 + 1 + ....& & %&(4.11) \\ \end{eqnarray}

is divergent because the partial sums {S}_{n} = n, which clearly diverge as n →∞.

Example 4.2: 

The geometric series

\begin{eqnarray} {\mathop{∑ }}_{k=0}^{∞}{x}^{k} = {1\over 1 − x},\quad |x| < 1.& & %&(4.12) \\ \end{eqnarray}

learn this one by heart.
This series converges for arbitrary (real or complex) numbers x with |x| < 1. [Please sketch the condition |z| < 1 for complex z as an an area in the complex plane.]

Proof of Eq. (4.33):

\begin{eqnarray} S& =& {\mathop{∑ }}_{k=0}^{∞}{x}^{k} = 1 + x + {x}^{2} + {x}^{3} + ... = 1 + x(1 + x + {x}^{2} + ...) = 1 + x ⋅ S ⇝ %& \\ S& =& {1\over 1 − x}. %&(4.13) \\ \end{eqnarray}

The problem with infinite series is that often it is not easy to decide if or if not they converge, e.g. for which values of x in the above example.

A necessary condition for convergence of S ={\mathop{ \mathop{∑ }}\nolimits }_{k=0}^{∞}{a}_{k} is that {a}_{k} → 0 as k →∞.

A sufficient condition for convergence: is the ratio test

Ratio test: Consider the series S :={\mathop{ \mathop{∑ }}\nolimits }_{k=0}^{∞}{a}_{k} and assume {a}_{k}\mathrel{≠}0 for all k > {k}_{0}. Define the ratio

\begin{eqnarray} R& :=& {\mathop{lim}}_{k→∞}\left |{{a}_{k+1}\over {a}_{k}} \right |⇝ %& \\ \array{ R < 1&\text{series is convergent}\cr R > 1&\text{series is divergent}.} & & %&(4.14)\\ \end{eqnarray}

For R = 1, the ratio test can’t decide whether the series is convergent or divergent.

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