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## No rational function

Real Analysis
Riemann
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### No rational function

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

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 $lim_{ n \to \infty} \frac{n^{m-k}}{lnn}=+ \infty$