# Show that for all Harmonic numbers H(k)H(k), the inequality H(2k)≤1+kH(2^k) \leq 1 + k holds for all natural numbers.

Hi so the question is prove that for all harmonic numbers H(k)=1+12+13+⋯+1kH(k) = 1 + \frac{1}{2} + \frac13 + \dotsb + \frac{1}{k} the inequality H(2k)≤1+kH(2^k) \leq 1 + k holds for all natural numbers. I’m not going to put my entire answer here but was wondering if someone could tell me if I’m on the right track.

So I got this far:
Say H(2k)=k+1H(2^k) = k + 1
and H(2k+1)=k+1+2k12k+1=k+1+12=k+32H(2^{k+1}) = k + 1 + 2^k \frac{1}{2^{k+1}} = k + 1 + \frac{1}{2} = k + \frac{3}{2}
and k+32≤1+(k+1)k + \frac{3}{2} \leq 1 + (k + 1)

Is this correct or at least am I on the right track??

=================

Please verify the edit and learn MathJax.
– Max
2 days ago

The inequality to prove does not seem to work for k=0k=0 and k=1k=1
– Claude Leibovici
2 days ago

@ClaudeLeibovici To which inequality you’re refering to? Since for k=0,1k=0,1 we have H(20)=H(1)=1≤1+0H(2^0) = H(1) = 1 \le 1 + 0 and H(21)=H(2)=32≤1+1=2H(2^1) = H(2) = \frac 32 \le 1 + 1 = 2
– Stefan4024
2 days ago

=================

3

=================

Let us prove something stronger through a powerful technique, creative telescoping.
We may recall that since in a neighbourhood of the origin we have
arctanh(x)=12log(1+x1−x)=x+x33+x55+x77+… \text{arctanh}(x) = \frac{1}{2}\log\left(\frac{1+x}{1-x}\right)=x+\frac{x^3}{3}+\frac{x^5}{5}+\frac{x^7}{7}+\ldots\tag{1}
the inequality
1n≤2arctanh(12n)=log(2n+12n−1)=1n+112n3+180n5+… \frac{1}{n}\leq 2\,\text{arctanh}\left(\frac{1}{2n}\right) = \log\left(\frac{2n+1}{2n-1}\right) = \frac{1}{n}+\frac{1}{12n^3}+\frac{1}{80n^5}+\ldots \tag{2}
surely holds for any n≥1n\geq 1. Additionally, log(2n+12n−1)\log\left(\frac{2n+1}{2n-1}\right) is a telescopic term, wonderful:
HN=N∑n=11n≤N∑n=1log(2n+12n−1)=log(2N+1)≤log(N)+1 \color{red}{H_N} = \sum_{n=1}^{N}\frac{1}{n}\color{red}{\leq} \sum_{n=1}^{N}\log\left(\frac{2n+1}{2n-1}\right) = \color{red}{\log(2N+1)}\leq \log(N)+1\tag{3}
In particular, H2k≤1+klog2H_{2^k}\leq 1+k\log 2.

The super-tight inequality
log(N+12)+γ≤HN≤log(N+12)+γ+124N(N+1) \log\left(N+\frac{1}{2}\right)+\gamma \leq H_N \leq \log\left(N+\frac{1}{2}\right)+\gamma+\frac{1}{24 N(N+1)} \tag{4}
where γ\gamma is the Euler-Mascheroni constant can be proved through the same technique.

– JeanMarie
2 days ago

Yes, you’re on the right track, but it seems that you’ve mixed something in the inductive step. In each “step” by going from H(2k)H(2^k) to H(2k+1)H(2^{k+1}) you add 2k2^k new numbers, each of which is less than 12k\frac{1}{2^k}, so the difference between H(2k)H(2^k) and H(2k+1)H(2^{k+1})is always going to be less than 11. Con you continue now?

\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}

By Induction !!!:

H2k+1=2 × 2k∑n=11n=2k∑n=11n+2 × 2k∑n=2k+11n<(1+k)+12k+1(2×2k−2k)=(1+k)+2k2k+1<1+(k+1)\begin{align} H_{2^{k + 1}} & = \sum_{n = 1}^{2\ \times\ 2^{k}}{1 \over n} = \sum_{n = 1}^{2^{k}}{1 \over n} + \sum_{n = 2^{k} + 1}^{2\ \times\ 2^{k}}{1 \over n} < \pars{1 + k} + {1 \over 2^{k} + 1}\pars{2 \times 2^{k} - 2^{k}} = \pars{1 + k} + {2^{k} \over 2^{k} + 1} \\[5mm] & < \bbx{1 + \pars{k + 1}} \end{align}