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 Post subject: No rational functionPosted: Fri Mar 10, 2017 10:37 am

Joined: Sat Nov 14, 2015 6:32 am
Posts: 159
Location: Melbourne, Australia
Prove that there exists no rational function such that

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

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

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 Post subject: Re: No rational functionPosted: 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 $lim_{ n \to \infty} \frac{n^{m-k}}{lnn}=+ \infty$

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