Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm doing some exercises and came across one that has two parts, as follows:

Given a transition matrix for a Markov Chain, $\mathbf{P}$, and a vector $\mathbf{f}$, $\mathbf{f}$ is harmonic if

$$ \mathbf{f} = \mathbf{P}\mathbf{f}$$

$(a)$ Show that if $\mathbf{f}$ is harmonic, then

$$ \mathbf{f}=\mathbf{P}^n\mathbf{f} $$

for all $n$

$(b)$ Using $(a)$, show that if $\mathbf{f}$ is harmonic,

$$ \mathbf{f} = \mathbf{P}^\infty \mathbf{f} $$

Am I incorrect in assuming that if $(a)$ holds, then $(b)$ holds by necessity? Are there any cases where proving that something holds for all $n$ does not prove that it holds as $n$ tends to infinity?

share|improve this question
3  
You need to refer back to the definition of $\textbf{P}^{\infty}$. –  Qiaochu Yuan Sep 6 '12 at 19:38
1  
If something holds for all $n$, then it clearly holds for $n$ as $n$ tends to $+\infty$. But you don't seem to be asking what happens as $n$ tends to $+\infty$, but instead what happens at $+\infty$. –  Hurkyl Sep 6 '12 at 20:22
    
@Hurkyl "If something holds for all $n$, then it clearly holds for $n$ as $n$ tends to $+∞$." What about the counterexample below? –  Pedro Tamaroff Sep 7 '12 at 1:47
    
@Peter: It's an example of a statement that holds for every term in any sequence of natural numbers tending towards $+\infty$, but whose (suitably interpreted) limit fails. –  Hurkyl Sep 7 '12 at 4:10
add comment

2 Answers

up vote 13 down vote accepted

Simple counterexample

$$\frac{1}{n}>0\text{ for all }n\in\Bbb N$$

share|improve this answer
add comment

The sum $$ \sum_{k = 1}^n \frac{1}{k} $$ is finite for all finite $n$, but $$ \sum_{k = 1}^\infty \frac{1}{k} $$

is infinite.

share|improve this answer
    
OK, let me be a nitpicker here, but what does it mean for a finite sum to be convergent? –  Pedro Tamaroff Dec 3 '12 at 22:59
    
@PeterTamaroff I agree it is an abuse of terminology. I meant only that the sum of finitely-many finite terms is finite. I've replaced my answer with something more concrete. –  Austin Mohr Dec 4 '12 at 0:37
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.