1

Does there exist a degree-2 polynomial with positive acceleration such that the real extension of the harmonic numbers surpasses it for all future values?

This was too big of a title and it's a really big mouthful, so I'll further elaborate on what I'm looking for.

$$ p(x) = ax^2 + bx + c \quad \quad a > 0, \quad (b, c) \in \mathbb R^2 $$

Above is the family of polynomials I care about. Positive acceleration (the highest degree term's coefficient), other coefficients just need to be real.

$$ H_k = \sum_{n = 1}^x \frac 1 n $$

Above is the definition of the $k$th Harmonic Number.

$$ H_x = H_k \quad \quad x \in \mathbb N $$

$H_x$ will be the real extension of $H_k$ to the reals.

$$ H_x = \int_0^1 \frac{1 - y^x}{1 - y} dy $$

My question is now, does there exist $p(x)$ as defined above, such that $p(x_0 + c) < H_{x_0 + c}$ for some $x_0$ and $c > 0$?

Axoren
  • 2,303
  • 1
    Another way to look at it. The "derivative" of harmonics, $H_{k+1}-H_k$, is $1/(k+1)$ which is always $<1$. But a polynomial $p(x)=ax^2+bx+c$ has "derivative" $p(x+1)-p(x)=2ax+b+a$. This grows without a bound, and is $>2$ for large enough $x$, say when $x>x_0$. Therefore whatever "early" lead the harmonics might have gotten (possibly with a suitable choice of $b$ and $c$) will quickly begin to diminish for $x>x_0$. At least by one, when you increase $x$ by one, so sooner rather than later the harmonics will lose. – Jyrki Lahtonen Jun 20 '16 at 11:39
  • @JyrkiLahtonen Yes, that was a bad copy of a formula. I'll correct it. – Axoren Jun 20 '16 at 19:37
  • @JyrkiLahtonen If you post your explanation as an answer, I'll accept it. It's sufficient. – Axoren Jun 20 '16 at 19:39

0 Answers0