Coursera Complex Analysis 6.1 Infinite Series

Aug 8, 2016

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

Slide 9 Example of absolute convergence inequality

<< to do >>

 

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