Is it possible to analytically solve the following sum for k: k * ∑(1/n), n = 1 to 2k? View the linked image to see the sum in question written in conventional notation: 
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K. Claesson
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When $k \geq 1$, it is known that $$\sum_{n=1}^{2k} \frac{1}{n}$$ is not an integer although I don’t know whether or not it will ever have a form so that multiplying by $k$ will yield an integer. – user328442 Nov 22 '17 at 14:28
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@user328442 , indeed, no number of that form can be an integer. There is some prime $p$ between $k$ and $2k$. That prime $p$ is in the denominator of our expression, and it cannot be canceled by multiplication with $k$. – G Tony Jacobs Nov 22 '17 at 15:23
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That's assuming $k$ is an integer. If $k$ is a half integer: $\frac{2k}{2}$, we still can't cancel $p$, because it is not a factor of $2k$. – G Tony Jacobs Nov 22 '17 at 15:25
2 Answers
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No.
The left hand side is strictly increasing and a quick computation has:
$$32 \sum_{n=1}^{64} \frac{1}{n} \approx 151.805$$ and
$$31 \sum_{n=1}^{62} \frac{1}{n} \approx 146.084$$
If we assume that $k$ need not be an integer then we must also check for the case where $k = \frac{63}{2}$.
$$\frac{63}{2} \sum_{n=1}^{63} \frac{1}{n} \approx 148.94$$
user328442
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1I think you can't assume that $k$ is an integer. I think you need also to check that for $k=63/2$ you get $\frac{63}{2}\sum_{n=1}^{63} 1/n<150$. – Jaakko Seppälä Nov 22 '17 at 14:41
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The sum $\displaystyle\sum_{n=1}^{2k}\frac1n$ approximately equals $\ln (2k)$, so we can get close to a solution by solving $k\ln(2k)=150$. By looking at a graph, this puts $k$ close to 35. Plugging in some numbers, we see that $35$ is too big:
$$35\displaystyle\sum_{n=1}^{70}\approx 169.149.$$
Adjusting, we see that even $32$ is too big:
$$32\displaystyle\sum_{n=1}^{64}\approx 151.805,$$
but $31$ is too small:
$$31\displaystyle\sum_{n=1}^{62}\approx 146.084.$$
I realize you asked for an analytic solution, but I wouldn't expect to find one. You can restate your question as $k\cdot H_{2k}=150$, where $H_j$ represents the $j$-th harmonic number, but it's not clear how that's helpful.
Where did this question arise?
G Tony Jacobs
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I am unsure on the exact origin of this problem. Some of my peers asked me to help them solve the equation. They're working on an experimental project concerning Zipf's law and the solution to this equation in combination with their experiment somehow speaks for the validity of Zipf's law. – K. Claesson Nov 22 '17 at 15:20
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If they need an exact answer, there isn't one, because the summation notation only makes sense when $2k$ is an integer. If a close one will do, then they might want $31$, or $32$, or $31.5$, depending on the context. – G Tony Jacobs Nov 22 '17 at 15:27