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I was playing around with adding fractions and noticed that when I start at any fraction of the form $1/n, n \in \mathbb{N}$, I get:

$$\frac{1}{3}+\frac{1}{4}\geq \frac{1}{2},\; \frac{1}{4}+\frac{1}{5}+\frac{1}{6}\geq\frac{1}{2}$$ After trying out for a few more numbers, I noticed that this always seems to happen. I am not incredibly surprised since $\sum\frac{1}{n}$ is divergent, but does anyone know of a proof for why $\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{k}\geq \frac{1}{2}$?

WaterDrop
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    How are you choosing $k$? Is the idea that $k$ is a function of $n$? Is the idea that $k$ is the least denominator where the sum of the reciprocals from $1/n$ to $1/k$ exceeds $1/2$? Is $k$ chosen some other way? For the two examples you gave, it is the case that each reciprocal on the left of the inequality is greater than the last reciprocal and the sum of that many copies of the last reciprocal is $1/2$... – Eric Towers Oct 03 '21 at 23:29
  • @EricTowers $k$ is the least denominator where the sum of the reciprocals exceeds $1/2$. – WaterDrop Oct 03 '21 at 23:32
  • ${1\over4}+{1\over5}$ is not greater than ${1\over2}$. If you add more than $k\over2$ terms greater than $1\over k$ then the inequality easily holds. – player100 Oct 03 '21 at 23:32
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    If "$k$ is the least denominator where the sum of the reciprocals exceeds $1/2$" then it is true by definition – Henry Oct 03 '21 at 23:33
  • We already know that $\sum\frac{1}{n}$ is divergent. So you are just subtracting a finite number (which is $\frac{1}{1} + \dots \frac{1}{n-1}$) from it, of course it will be divergent again, and it will exceed $\frac{1}{2}$ and any real number, indeed. – Ali Dursun Oct 03 '21 at 23:35
  • @Henry I think I understand now. Perhaps I worded it incorrectly. What if I wanted to show that there exists a $k>n$ such that the summation of $1/n + 1/(n+1) + ... + 1/k$ is greater or equal to 2? – WaterDrop Oct 03 '21 at 23:36
  • For $\frac12$ try any $k >1.649n-1$. For $2$ try any $k > 7.39n-1$ – Henry Oct 03 '21 at 23:55

3 Answers3

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Is this what you are interested in?

\begin{align*} \frac{1}{n + 0} + \frac{1}{n+1} + \frac{1}{n+2} + \ldots + \frac{1}{2n} & > \frac{1}{n + n} + \frac{1}{n + n} + \frac{1}{n + n} + \ldots + \frac{1}{2n}\\\\ & = \frac{n + 1}{2n}\\\\ & = \frac{1}{2} + \frac{1}{2n}\\\\ & > \frac{1}{2} \end{align*}

Átila Correia
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I'm not sure what $k$ is, but you seem to be asking if

$$\sum_{i=n}^{2n-2}\frac{1}{i}\geq \frac{1}{2}$$

for $n\geq 2$. We have that

$$\sum_{i=n}^{2n-2}\frac{1}{i}=\sum_{i=1}^{2n-2}\frac{1}{i}-\sum_{i=1}^{n-1}\frac{1}{i}=H_{2n-2}-H_{n-1}$$

where $H_n$ is the $n$th harmonic number. A bound for the harmonic numbers is

$$\ln(n)\leq H_n-\gamma\leq \ln(n)+\frac{1}{2n}$$

This implies that

$$H_{2n-2}-H_{n-1}\geq \ln(2n-2)-\ln(n-1)-\frac{1}{2n}=\ln(2)-\frac{1}{2n}>\frac{3}{5}-\frac{1}{2n}$$

For $n\geq 5$ this is

$$\frac{3}{5}-\frac{1}{2n}\geq \frac{3}{5}-\frac{1}{10}=\frac{1}{2}$$

For $2\leq n\leq 4$ the proposition can be manually checked.

QC_QAOA
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Let's consider $$\frac{1}{m}=\int\limits_m^{m+1} \frac1m \,dx \gt \int\limits_m^{m+1} \frac1x \,dx = \log(m+1)-\log(m)$$

So similarly $$\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{k} \gt \int\limits_n^{k+1} \frac1x \,dx =\log(k+1)-\log(n)$$ which will always exceed $q$ when $$k \ge e^q n-1$$ and may exceed it for slightly smaller $k$. Note when $q=\frac12$ you have $e^{1/2} \approx 1.649$

Henry
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