It is currently Wed May 23, 2018 9:37 pm


All times are UTC [ DST ]




Post new topic Reply to topic  [ 2 posts ] 
Author Message
 Post subject: No rational function
PostPosted: Fri Mar 10, 2017 10:37 am 

Joined: Sat Nov 14, 2015 6:32 am
Posts: 139
Location: Melbourne, Australia
Prove that there exists no rational function such that

$$ f(n)=1+ \frac{1}{2} + \cdots + \frac{1}{n} \quad \text{forall} \; n \in \mathbb{N} $$

_________________
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$


Top
Offline Profile  
Reply with quote  

 Post subject: Re: No rational function
PostPosted: Sat Jul 29, 2017 1:06 pm 

Joined: Tue Nov 24, 2015 7:47 pm
Posts: 13
Suppose there exists a rational function ( i assume you mean the function is a quotient of two polynomials ) .

Let $ f(x)=\frac{P(x)}{Q(x)} $
Clearly since the harmonic series diverges $ lim_{ n \to \infty} f(n)=+\infty $

that means that $ deg(P(x)) > deg(Q(x)) $


Since $ lim_{ n \to \infty} \frac{H_n}{lnn} \stackrel{Cezaro-Stolz}{=} lim_{ n \to \infty} \frac{1}{(n+1)ln(\frac{n+1}{n})}=1 $

Therefore $ lim_{ n \to \infty} \frac{f(n)}{lnn}=1 $
But $$ lim_{ n \to \infty} \frac{1}{lnn} \frac{a_m n^m+...+a_0}{b_k n^k+...b_0} \Rightarrow
lim_{ n \to \infty} \frac{f(n)}{lnn} = lim_{ n \to \infty} \frac{1}{lnn}n^{m-k} \frac{a_m+....+\frac{a_0}{n^m}}{b_k+...+\frac{b_0}{n^k}}$$

Which leads to a contradiction because $ lim_{ n \to \infty} \frac{n^{m-k}}{lnn}=+ \infty $


Top
Offline Profile  
Reply with quote  

Display posts from previous:  Sort by  
Post new topic Reply to topic  [ 2 posts ] 

All times are UTC [ DST ]


Mathimatikoi Online

Users browsing this forum: No registered users and 1 guest


You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot post attachments in this forum

Search for:
Jump to:  
Powered by phpBB® Forum Software © phpBB Group Color scheme created with Colorize It.
Theme created StylerBB.net