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I came across this problem whilst exploring the asymptotic behaviour (or not) of different generalised harmonic numbers. I am interested in the point of 'cross-over' between a generalised harmonic number where the denominator of the summand is raised to a power, and a non-exponential harmonic sum operating on some subset of the natural numbers.

For example, take the generalised harmonic number $H_x^{(k)}=\sum_{n=1}^x \frac{1}{n^k}$, and a harmonic number operating only on odd denominators $G_x=\sum_{n=1}^x \frac{1}{2 n-1}$.

Clearly, there exist values of $x,k$ such that $G_x<H_x^{(k)}$ and values such that $H_x^{(k)}<G_x$. Thus there exists a value $c=G_{x_0}$ such that

$$G_{x_0}=c<H_{x_0}^{(k)}=\sum_{n=1}^x \frac{1}{n^k}$$ and $$H_{{x_0}+2}^{(k)}<G_{{x_0}+2}=c+\frac{1}{2x_0+1}+\frac{1}{2x_0+3}$$ or $$H_{{x_0}+2}^{(k)}-c<\frac{1}{2x_0+1}+\frac{1}{2x_0+3}$$

The values of $c,x_0,k$ are obviously co-dependent. I am searching for a way to solve for $x_0$ or at least put bounds on it.

I am interested in how to approach this algebraically rather than numerically. This is a single simple example of $G$ and I want to be able to explore how to solve such problems generally, for whatever pattern of $G$ I choose (provided it's formulable!).

Algebraically, how do I put bounds on $x_0$ in terms of $c,k$?

(I am an amateur, so I need a fair amount of hand-holding, hence the bounty.)

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  • $\begingroup$ Hi @Saad. Humble apologies - please see revision. $\endgroup$ Commented May 4, 2020 at 13:01
  • $\begingroup$ It does. My bad - a hangover from the over-terse prec=vious question, now fixed. $\endgroup$ Commented May 4, 2020 at 16:31

1 Answer 1

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I am interested in how to approach this algebraically rather than numerically. This is a single simple example of $G$ and I want to be able to explore how to solve such problems generally, for whatever pattern of $G$ I choose (provided it's formulable!).

We can approximate the sums by the integrals and then to deal with the resulting functions. For instance, for $k<1$, $$H^{(k)}(x)\approx \int_{1}^{x} \frac 1{t^k} dt=\frac{t^{1-k}}{1-k}{\Huge |}^{x}_1=\frac{1}{1-k}\left(x^{1-k}-1\right).$$ For $k=1$ and natural $x$, according to Wikipedia, $$H^{(1)}(x)\sim\ln x+\gamma+\frac{1}{2x}- \frac{1}{12x^2}+\frac{1}{120x^4}-\dots,$$
where $\gamma\approx 0.5772156649$ is the Euler–Mascheroni constant. For $k>1$ when $x$ tends to infinity, the sequence $H^{(k)}_x$ converges to Riemann zeta function $\zeta(k)$.

Similarly we have $$G(x)\approx \int_1^{x}\frac {1}{2t-1}dt=\frac 12\ln (2x-1).$$

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    $\begingroup$ Hi @Alex. How would I isolate the required value of $x$ from this? $\endgroup$ Commented May 6, 2020 at 10:52
  • $\begingroup$ @RichardBurke-Ward A good question. I have to confess that I didn’t understand the details of you question. But I guess that an analytic approach for bounds for $x_0$ should start from the above observations. Maybe somebody more clever than me would be able to finish it. $\endgroup$ Commented May 6, 2020 at 17:31
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    $\begingroup$ Since no one else appears to have anything to add, I'm going to mark your answer as definitive. If there is anything else you could add, of course I'd be grateful. $\endgroup$ Commented May 9, 2020 at 8:31

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