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No rational function

Real Analysis
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Riemann
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No rational function

#1

Post by Riemann » Fri Mar 10, 2017 10:37 am

Prove that there exists no rational function such that

$$ f(n)=1+ \frac{1}{2} + \cdots + \frac{1}{n} \quad \text{forall} \; n \in \mathbb{N} $$
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
dr.tasos
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Re: No rational function

#2

Post by dr.tasos » Sat Jul 29, 2017 1:06 pm

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 $
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