$\frac{1}{1000}(H_n) \geq 1$ 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.

-
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

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 http://en.wikipedia.org/wiki/Harmonic_number
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 http://numbers.computation.free.fr/Constants/Gamma/gamma.pdf . 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.
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[{$MaxExtraPrecision = 2000}, HarmonicNumber[#] > 1000 & /@ ({#, # - 1} &[Ceiling[Exp[1000 - EulerGamma]]])]  - read the question again. It's the harmonic series, that insanely huge number that$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