## Slide 2 Definition

$\Sigma_{k=0}^\infty {a_k} = a_0 + a_1 +a_2 + ... + a_n + ...$

which converges if sequence of partial sums, Sn, converges

## Slide 3 Example

$\Sigma_{k=0}^\infty {z^k}$

$S_n = 1 + z + z^2+ ... z^n$

Formula in usual way $S_n = \frac{1-z^{n+1}}{1-z}$

Since $z^{n+1} \to 0 \textrm{ as } n \to \infty$ as long as |z| < 1

then $\Sigma_{k=0}^\infty {z^k} = \frac{1}{1-z}$ for |z| < 1.

## Slide 4 Convergence and Divergence

Theorem: If a series converges, then $a_k \to 0 \textrm{ as k } \to \infty$ (*)

The converse (swap propositions), if $a_k \to 0$ then sum converges is not necessarily true – classic example is the harmonic series.

Contrapositive (*) $a_k \to \infty \textrm{ or constant as k } \to \infty$ then series diverges, of course is true.

For 1/(1 – z) , this diverges when |z| >= 1

## Slide 5 Real and Imaginary parts of series

z = $re^{i \theta}$ so  $z^k \; = r^ke^{ik \theta} = r^kcos\theta \; + \; i r^k sin \theta$

So $\Sigma_{k=0}^\infty {z^k} = \Sigma_{k=0}^\infty r^kcos\theta \; + \; i \Sigma_{k=0}^\infty r^k sin \theta$

<< more , separate real and imaginary parts>>

## Slide 6 Another example of convergence. Part I

Does $\Sigma_{k=1}^\infty \frac{i^k}{k}$ converge?

If take modulus of the terms, $|\frac{i^k}{k}|$ then get harmonic series which does not converge.

What about original series on this slide? Split into real and imaginary parts.

## Slide 7 Another example of convergence. Part II

Preliminaries: k is even, $i^k = i^{2n} = (-1)^n$ ( a real number)

Similarly, k is odd, $i^k = i^{2n+1} = i(-1)^n$ (a purely imaginary number)

So $\Sigma_{k=1}^\infty \frac{i^k}{k} = \Sigma_{n=1}^\infty \frac{i^{2n}}{2n} + \Sigma_{n=0}^\infty \frac{i^{2n+1}}{2n+1}$ (note the limits).

Using prelim results and simplifying: $\frac{1}{2}\Sigma_{n=1}^\infty \frac{-1^{n}}{n} + i \Sigma_{n=0}^\infty \frac{-1^{n}}{2n+1}$

The first sum is the alternating harmonic series which converges (can justify by looking at intervals on real line)

and (probably) because denominators of terms second series getting smaller and series is alternating this also converges.

## Slide 8 Absolute Convergence

Definition ? $\Sigma_{k=0}^{\infty}a_k$ converges absolutely if the series $\Sigma_{k=0}^{\infty}|a_k|$ converges

<< examples >>

If $\Sigma_{k=0}^{\infty}a_k$ converges absolutely then it converges and $|\Sigma_{k=0}^{\infty}a_k| \leq \Sigma_{k=0}^{\infty}|a_k|$ converges absolutely

<< to do >>