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With the $n$th harmonic number defined as $$ H_n = \sum\limits_{k = 1}^n {\frac{1}{k}} $$ I'm supposed to find the minimum EXACT value of $n$ such that $H_n>100$ . I could only find approximations, but my professor demands an exact value . Is this possible ? If so, how can we find it?

artmath
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  • Hint: Logarithms. – user_of_math Dec 06 '14 at 17:59
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    Your notation here is not just sloppy but outright wrong. The correct definition is $H_n = \sum_{k=1}^n \frac{1}{k}$. – Ian Dec 06 '14 at 17:59
  • Sorry, just fixed it. Thank you for pointing it out – artmath Dec 06 '14 at 18:01
  • I guess you get approximations for $H_n$, I suppose $n$ is integer. If (let me invent the values) $H_{31}=99.8738...$ and $H_{32}=100.0235...$, what your professor wants is 32 – Octania Dec 06 '14 at 18:04
  • @Octania I don't think artmath is confused about that part. I'd guess the confusion lies in how to find $n$ algebraically without guessing and checking – graydad Dec 06 '14 at 18:05
  • @graydad, exactly ! I'm really curious if there's a way to find the value of $n$ by hand, using mathematical techniques, rather than calculators – artmath Dec 06 '14 at 18:08
  • @artmath I have an idea to solve this without using integrals. If you give me a minute I'll post what I have! – graydad Dec 06 '14 at 18:09
  • Really looking forward to seeing your idea ! – artmath Dec 06 '14 at 18:11
  • After digging into this problem I see that it is more complex than I had originally anticipated. Looks like you have some good answers from more experienced users on here. There's also this wikipedia page that casts your problem in terms of a differential equation. http://en.wikipedia.org/wiki/Ant_on_a_rubber_rope And the exact answer to your question can be found here http://mathworld.wolfram.com/HarmonicSeries.html – graydad Dec 06 '14 at 19:03

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We have: $$ \log(n+1/2)+\frac{1}{24n^2}-\frac{1}{24n^3}\leq H_n-\gamma \leq \log(n+1/2)+\frac{1}{24n^2}$$ hence the smallest natural number for which $H_n>100$ is about $m=\left\lfloor e^{100-\gamma}-\frac{1}{2}\right\rfloor.$

Notice that $\psi(x)=\frac{\Gamma'(x)}{\Gamma(x)}=-\gamma+H_{x-1}$ is a concave function, since: $$ \psi'(x) = \sum_{n\geq 0}\frac{1}{(n+x)^2}.$$ This gives that the Newton's method with starting point $m$ is a really wise choice to find the exact value of the least integer $n$ such that $H_n>100$.

Jack D'Aurizio
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    Your estimate provides an answer of about [1.50927... x 10^{43}], which is very close to @DumpsterDoofus's answer. – user_of_math Dec 06 '14 at 18:25
  • @Jack D'Aurizio, you deserve the bounty undoubtedly, but I'm interested in a more detailed answer shaped in the form of a proof, if you have the time. – Victor Dec 25 '14 at 23:24
  • I must agree, the estimate provides a lot of insight, but is there any way to hit this precise value $n=15092688622113788323693563264538101449859497$ through mathematical technique alone without involving any computation software whatsoever ? – Victor Dec 26 '14 at 02:42
  • @Victor: through some version of Stirling approximation is quite easy to recover the first, let us say, six or seven significant figures, but the exact value of such huge number is just a problem in approximation theory, and I think there is no way to avoid some root-finding algorithm. This problem is not much different from computing $\lfloor e^{1000}\rfloor$. Can you do it by hand? – Jack D'Aurizio Dec 26 '14 at 08:58
  • Pardon me for my sloppy way of expressing my interest . Indeed, computation by hand of such a value might not only be tedious, but impossible. What I'm concerned with is finding a closed form for $n$'s representation. Namely, say if one manages to find $n=e^{100}$ that would be sufficient for my purposes, I wouldn't be interested in an explicit numerical form – Victor Dec 27 '14 at 02:51
  • @Victor: honestly, I have no idea if a nice closed form representation for $n$ (or its logarithm $99.42278433509846713939348791\ldots$) exists. My bet is not, since there is nothing special in $100$ as a point in which to evaluate an inverse function. – Jack D'Aurizio Dec 27 '14 at 09:35
  • Obviously we can exploit the machinery of Lagrange inversion formula, giving a series whose coefficients depend on $\psi^{(n)}(100)$, but this is not a "nice" closed form. – Jack D'Aurizio Dec 27 '14 at 09:42
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The simplest pertinent estimate is based on Riemann sums. Specifically, if the step size is $h=1$, then $H_n$ is the corresponding upper Riemann sum for

$$\int_1^n \frac{1}{x} dx.$$

Hence $H_n \geq \ln(n)-\ln(1) = \ln(n)$. Solving $\ln(n) \geq 100$ we have $n \geq e^{100}$.

For the other bound, we have that

$$H_n - 1 \leq \int_1^n \frac{1}{x} dx \Rightarrow H_n \leq \ln(n)+1.$$

Solving $\ln(n)+1 \leq 100$ we have $n \leq e^{99}$, so your minimal answer will be between $e^{99}$ and $e^{100}$.

This is clearly quite crude. However, about the only way to improve this is to accurately estimate the Euler-Mascheroni constant $\gamma$, since when $n$ is large we have $H_n \approx \ln(n) + \gamma$, meaning that your minimal answer should be in the vicinity of $e^{100-\gamma}$.

Ian
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    There is a massive guess-and-check, since there are many integers between $e^{100}$ and $e^{100-\gamma}$. – Jack D'Aurizio Dec 06 '14 at 18:10
  • Yes, but merely 150 or so iterations of bisection will find the answer for you. :) – John Hughes Dec 06 '14 at 18:25
  • @JohnHughes Precomputing $H_{e^{99}}$ and $H_{e^{100}}$ to even prepare for bisection would be a terrible task. – Ian Dec 06 '14 at 18:27
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    I completely agree...I was being whimsical rather than serious. Clearly Jack D'Aurizio's answer (or a generalization using higher order approximations) is the way to go. – John Hughes Dec 06 '14 at 18:28
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The answer is $$\lceil\chi\rceil$$ where $\chi$ is the solution of $$\text{N}(\chi)=100$$ where $\text{N}$ is the analytic continuation of the harmonic sum function, and $\lceil\rceil$ is the integer ceiling function.

This can be quickly calculated in Mathematica:

Ceiling[x] /. Solve[HarmonicNumber[x] == 100, x][[1]]

which instantly yields the answer:

15092688622113788323693563264538101449859497
DumpsterDoofus
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    This answer doesn't clarify any of the underlying mathematics behind the question; it just feeds it to Mathematica. It'd probably be better to leave this as a comment on the question. – Milo Brandt Dec 06 '14 at 18:26
  • @Meelo: On the contrary, the underlying mathematics to the question is simply that of executing a root-finding procedure for a continuous extension of a discrete strictly-increasing sum, and nothing more. The "high-school" version of the answer uses the obvious Riemann estimation method, which is, as John Hughes pointed out, completely useless. – DumpsterDoofus Dec 06 '14 at 18:59
  • @Meelo: Meanwhile, Jack's method of connecting the analytic continuation to the polygamma function family gives a far more computationally more tractable method (and arguably a deeper connection to the realm of special functions), but in the end, all methods do not give explicit integer solutions, so what is the point of ruling out an answer which gives the integer solution in one step? – DumpsterDoofus Dec 06 '14 at 19:00
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    The integer solution could be useful to the problem, which is why I suggest it as a comment. However, the question is seeking to know how to quickly solve something like your HarmonicNumber[x]==100 - and that all takes place within Mathematica. Given that we don't know the algorithm within Mathematica, this merely answer the question by asking, "How does Mathematica solve the question?" – Milo Brandt Dec 06 '14 at 19:18
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    Don't understand what people have against plain results. Of course we have CAS tools at our disposal, and anybody can press a button. So what? Therefore (+1) for this answer. – Han de Bruijn Dec 27 '14 at 16:22
  • this is just a way of solving the problem using brute force. – 420 Mar 13 '18 at 09:20