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$\frac{1}{1000}(H_n) \geq 1$ where $H_n$ represents a partial sum of the harmonic series, starting at $1$ and ending at $\frac{1}{n}$. What I want is a method of finding the smallest value that $n$ could possibly be. This is in relation to a puzzle, and the author says the smallest value is $e^{1000 - y} + 1$ or $e^{1000 - y} - 1$ where $y$ is Euler's constant. He does not explain how he got this answer. I would like to know how he got this answer, and how to solve problems like it. Feel free to explain in any way you like, and thank you in advance.

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Note: any suggestions for tags would be appreciated. I didn't find any that met the needs of this question. – recursive recursion Feb 1 '14 at 17:57
up vote 3 down vote accepted

I think the author of your problem may be using the asymptotic expansion $$ H_n \sim \ln n + \gamma + \frac{1}{2n} - \frac{1}{2n^2} + \frac{1}{120n^4} - \cdots $$ where $\gamma = 0.577215664\ldots$ is the Euler–Mascheroni constant and $f(n) \sim g(n)$ means $\displaystyle{\lim_{n \rightarrow \infty} \dfrac{f(n)}{g(n)} = 1}$. This can be found on the wikipedia page

You may find more useful the explicit error bound $$ \frac{1}{2n+1} < H_n - \ln n - \gamma < \frac{1}{2n} $$ which can be found on page 2 of the paper . There are better error bounds in the paper as well. Follow the references given in the paper for this bound and others like it to see how they are proved.

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Great explanation. I did not know about the asymptotic expansion you mentioned. Thank you! – recursive recursion Feb 2 '14 at 15:44

That can't be right; if $n = e^{100000-\gamma}+1$, $H_n$ would be approximately $100000$, not $1000$ as you require. This is due to the relationship $$\lim_{n \to \infty} H_n - \log n = \gamma,$$ where $\gamma \approx 0.57721566490153286061\ldots$ is Euler's constant. Instead, the value should be $n = \lceil e^{1000-\gamma} \rceil$. The precise solution is $$\begin{align*} n &= 11061151102660493564107470558442113839302800185257 \\ &\quad 73739364709523772183545751724012754575975790447298 \\ &\quad 73152469512963401398362087144972181770571895264066 \\ &\quad 11408896818235684297782376446217982198174444873178 \\ &\quad 54086291163219199578560346058778552126670922875201 \\ &\quad 05386027668843119590555646814038787297694678647529 \\ &\quad 53371876940106926942747586879353194469643569674555 \\ &\quad 92893266101322085042577214698292107044628765749153 \\ &\quad 62273129090049477919400226313586034. \end{align*}$$ This was verified in Mathematica (v.9) with the command

Block[{$MaxExtraPrecision = 2000}, HarmonicNumber[#] > 1000 & 
       /@ ({#, # - 1} &[Ceiling[Exp[1000 - EulerGamma]]])]
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read the question again. It's the harmonic series, that insanely huge number that $n$ is would be put over one. – recursive recursion Feb 2 '14 at 3:54
I have not made a mistake. You wrote $\frac{1}{1000} H_n \ge 1$, which is equivalent to finding $H_n \ge 1000$. Because $H_n \sim \log n + \gamma$, it immediately follows that $n \approx e^{1000 - \gamma}$, not $e^{100000 - \gamma}$ as you had written. Either you made a typo, or your reasoning is incorrect. – heropup Feb 2 '14 at 5:33
You're right, it was a typo. Let me fix that... – recursive recursion Feb 2 '14 at 15:41

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