No rational function
No rational function
Prove that there exists no rational function such that
$$ f(n)=1+ \frac{1}{2} + \cdots + \frac{1}{n} \quad \text{forall} \; n \in \mathbb{N} $$
$$ f(n)=1+ \frac{1}{2} + \cdots + \frac{1}{n} \quad \text{forall} \; n \in \mathbb{N} $$
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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 $
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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