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I am trying to find the smallest $n$ such that $1+1/2+1/3+....+1/n \geq 9$,

I wrote a program in C++ and found that the smallest $n$ is $4550$.

Is there any mathematical method to solve this inequality.

5 Answers5

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What you are looking for is $H_n = \sum_{1 \le k \le n} k^{-1}$. It is known that $H_n \approx \ln n + \gamma - \dfrac{1}{12 n^2} + \dfrac{1}{120 n^4} - \dfrac{1}{252 n^6} + O\left(\dfrac{1}{n^8}\right)$, where $\gamma \approx 0.5772156649$ is Euler's constant.

vonbrand
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  • ... $H_n$ is the sum you are computing. – liamdalton Apr 27 '13 at 19:52
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    This nails it: $e^{9-0.5772} \approx 4549.62$. – lhf Apr 27 '13 at 21:17
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    @julien Right, copied the signs wrong :-( In any case, the approximation without the series already gives enough to solve your question. This is an asymptotic series the error is less than the first omited term. – vonbrand Apr 27 '13 at 21:49
  • @julien, note that $B_{2 n} \sim (-1)^n \left( \dfrac{n}{\pi e} \right)^{2 n}$, so the series does not converge. It can be shown that the error is less than the first term left out. – vonbrand Apr 28 '13 at 13:16
  • Right, thanks. I found that on wikipedia afterwards. Can it really be shown that the error is less than the first term left out at every order? Will Jagy, who knows this formula pretty well, says this is for sure only for very little orders. And yes, it can be shown then. And even more. We have inequalities. That is the point of my answer, inspired by Will Jagy. And this is why it is not sufficient to just give an asymptotic expansion. The more difficult part comes after that. – Julien Apr 28 '13 at 13:32
  • @1015 If a series is an alternating, convergent series, then the error does not exceed the absolute value of the first term left out. – Steven Alexis Gregory Jan 19 '18 at 18:24
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We have $$\sum_{k=1}^N \dfrac1k \sim \log(N) + \gamma + \dfrac1{N} - \dfrac1{12N^2} + \mathcal{O}(1/N^3)$$ We want the above to be equal to $9$, i.e., $$9 = \log(N) + \gamma + \dfrac1{N} - \dfrac1{12N^2} + \underbrace{\mathcal{O}(1/N^3)}_{\text{can be bounded by }1/N^3}$$ We have $\log(N) + \gamma < 9 \implies N \leq 4550$. Also, $\log(N) + \gamma +\dfrac1{N} > 9 \implies N \geq 4550$. Hence, $$N = 4550$$

Here is the asymptotic from Euler–Maclaurin. We have $$\sum_{k=a}^b f(k) \sim \int_a^b f(x)dx + \dfrac{f(a) + f(b)}2 + \sum_{k=1}^{\infty} \dfrac{B_{2k}}{(2k)!}\left(f^{(2k-1)}(b)-f^{(2k-1)}(a)\right)$$ In our case, we have $f(x) = \dfrac1x$, $a=1$ and $b=N$. This gives us $$\sum_{k=1}^N \dfrac1k \sim \log(N) + \gamma + \dfrac{1}{2N} - \sum_{k=1}^{\infty} \dfrac{B_{2k}}{2k}\dfrac1{N^{2k}}$$

  • Very nice, thanks a lot. –  Apr 27 '13 at 19:54
  • What if $O(1/N^3)=10^{10^{10}}/N^3$? I know it is not... – Julien Apr 27 '13 at 20:03
  • @julien Valid point. But as you said, it is not. –  Apr 27 '13 at 20:12
  • @julien, in an asymptotic expansion, your accuracy at any fixed truncation stays pretty good. You should truncate at a later point, depending on the size of the argument, to get maximum accuracy. – Will Jagy Apr 27 '13 at 20:16
  • @julien, meanwhile, this has been asked a million times. See http://en.wikipedia.org/wiki/Harmonic_number#Calculation The standard references are Abramowitz and Stegun, available from Dover, and Gradshtyn and Ryzhik. There are also book about asymptotic expansions with proofs. But this result is quite old. And long. – Will Jagy Apr 27 '13 at 20:21
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    @WillJagy I am aware of the asymptotic exapnsion of the harmonic series...I am afraid you don't see my point. The problem is to know whether the remainder is $\leq \frac{1}{N^k}$ or $\leq \frac{10^{10^{10^{10}}}}{N^k}$. In the latter case, how do you find the first $n$ that makes your sequence greater than $9$ just looking at the truncation? – Julien Apr 27 '13 at 20:25
  • @julien Have added the asymptotic. Should be able to bound the error using this. –  Apr 27 '13 at 20:28
  • @Potato Yes. Thats why I wrote especially that we can bound it by $1/n^3$. –  Apr 27 '13 at 20:30
  • @julien, given your strong feelings on how much detail is necessary, I think you should post your own answer. – Will Jagy Apr 27 '13 at 20:50
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    @WillJagy Do you really not see my point? I did not mean to bother anyone... – Julien Apr 27 '13 at 20:52
  • @julien, the way to get explicit bounds is in my answer at http://math.stackexchange.com/questions/306371/simple-proof-of-showing-the-harmonic-number-h-n-theta-log-n where, for this problem you replace $\log(n+1)$ by the first few terms of the Bernoulli number expansion, show one version is low and increases, one version is high and decreases, then the common limit is $\gamma.$ How about if you do that? – Will Jagy Apr 27 '13 at 21:27
  • @WillJagy This is great, I was not aware of this at all. So $\ln n+\gamma<H_n<\ln(n+1)+\gamma$. This proves that $H_{4548}<9$ and $H_{4550}>9$. We're almost there! It is either $4549$ or $4550$... – Julien Apr 27 '13 at 21:34
  • @julien, for this problem it suffices to show that the version truncated but including $-1/12n^2$ starts low but increases, while the version including $1/120n^4$ starts high but decreases. I need to go to the drugstore. i will probably fiddle with this when I get back, if I see no answer by you. The series for $\log(1+x)$ is strictly alternating and so, for $0 \leq x < 1$ gives easy explicit bounds, in this case $x=1/n.$ – Will Jagy Apr 27 '13 at 21:35
  • @WillJagy I see it. Of course, I'll let you mention your argument when you have time. Regards. – Julien Apr 27 '13 at 21:40
  • @julien, I already know how to do this, and I don't need points. I really, really want you to learn this, by doing a very specific problem, the only real difficulty being messy common denominators, which you can check in a CAS if you like. If you are uncomfortable accepting points, make it Community Wiki. – Will Jagy Apr 27 '13 at 21:46
  • It seems, given the equivalent of $B_{2n}$ found on wikipedia, that the general term of the series coming from Euler-MacLaurin does not converge. So the formula you wrote at the end ould not hold. Is this right? – Julien Apr 28 '13 at 13:44
  • @julien Nope. This is a typical asymptotic series. Cut it off at a certain $k$, and as $N \to \infty$ it gets more accurate. However, the series is not convergent in $k$. I have replaced the $=$ sign by $\sim$ sign now to convey this. –  Apr 28 '13 at 17:22
  • I pointed that out because you wrote down the $k$ series as if it converged. Also, the problem here is not to let $N$ tend to $\infty$, but rather to have an accurate estimate from $N=1$. And if your O is indeed less than $1/N^3$, your expansion works well. I'm just not sure how to get this estimate. – Julien Apr 28 '13 at 17:27
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The idea was given to me by Will Jagy. I am just writing it down because he suggested I do so.

Let $H_n:=\sum_{k=1}^n\frac{1}{k}$ the $n$-th harmonic number. It is known (see here or user17762's answer above) that $$ H_n=\ln n+\gamma+\frac{1}{2n}+O\left( \frac{1}{n^2}\right) $$ where $\gamma$ denotes Euler-Mascheroni constant. It is not sufficient to know that to answer your question. You need inequalities, or a precise control of the error term in the expansion. Here are two inequalities which suffice.

Claim: we have $$ S_n:=\ln n+\gamma <H_n< \ln n+\gamma+\frac{1}{2n}=:T_n $$ for all $n\geq 1$.

Application: with the lhs, a calculator, and $\gamma\simeq 0.5772156649$, we find $H_{4550}>9.00009817>9$. With the rhs, we get $H_{4549}<8.99998828<9$. So $n=4550$ is indeed the first such $n$.

Proof of the claim: first, let us show that $x_n:=H_n-S_n$ decreases. Indeed $$ x_n-x_{n+1}=-\ln n+\ln(n+1)-\frac{1}{n+1}=-\ln\left(1-\frac{1}{n+1} \right)-\frac{1}{n+1}>0 $$ where the last inequality follows for instance from the study of $f(x)=-\ln(1-x)-x$ on $(0,1)$. Since $\lim x_n=0$, it follows that $x_n>0$ for all $n\geq 1$. That is the lhs inequality.

Now let $y_n:=H_n-T_n$ and let us check that it increases. We have $$ y_{n+1}-y_n=\frac{1}{2(n+1)}+\frac{1}{2n}-\ln\left(1+\frac{1}{n}\right)>0 $$ where the inequality follows from the fact that the corresponding function of $x$, instead of $n$, is decreasing on $(0,+\infty)$ with limit $0$ at $+\infty$. So $y_n$ is increasing with limit $0$. This means that $y_n<0$ for all $n\geq 1$, which proves the rhs inequality. QED.

Julien
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  • @WillJagy Here it is. Actually, I took a lazier path which suffices. Please let me know if there is anything wrong. Thanks a lot for showing me these inequalities. – Julien Apr 27 '13 at 23:42
  • That's right. The fact that this is decisive for threshold 9 is somewhat a matter of luck. The usual practice is to get the size of the window an extra order smaller then the last term added, that being $1/n$. In this case, that would mean considering your $T_n$ and $U_n = T_n - \frac{1}{12 n^2}.$ For example, what you have written does not suffice to find the first $H_n$ that exceeds 2, or exceeds 3, or 5, or 12. – Will Jagy Apr 28 '13 at 01:37
  • @WillJagy Yes. I started doing it with the order $2$ on the left and $4$ on the right. But I took a little break in doing so, which was more cumbersome, to try some computations. And I was surprised to see that these easier estimates suffice. – Julien Apr 28 '13 at 02:14
  • Well, if you do one more step, (and I recommend just the one additional) you will find that, as is common for strictly alternating asymptotic series, the true value is bounded between consecutive terms in the asymptotic series. This is not guaranteed when all you have is an asymptotic series, but is guaranteed for strictly alternating convergent infinite series. So you see, I was never worried about the issue that bothered you. – Will Jagy Apr 28 '13 at 02:46
  • @WillJagy As I don't know Bernoulli numbers very well, I don't know that $\frac{|B_{2k}|}{2kN^{2k}}$ is nonincreasing. Otherwise, I would not have worried about that either as I know Leibniz criterion and its consequences. Is it really straightforward to see that the latter is nonincreasing? – Julien Apr 28 '13 at 02:50
  • I have no idea. I'm just showing you how to find the threshold value for 12 instead of 9, emphasizing the aspect of asymptotic series that you appear to find interesting. Oh, I'm not even sure the alternation of the Bernoulli numbers continues after 1/6, -1/30, 1/42, -1/30, 5/66, -691/2730,7/6. – Will Jagy Apr 28 '13 at 03:06
  • @WillJagy I got your point about $12$, and the required inequalities can be proved with the same approach. I'm just wondering if the fact that $H_n$ can be bounded above and below, up top any order (this is what you seem to suggest, no?) by the even and the odd partial sums of the alternating series isn't simply due to the fact that this series satisfies the Leibniz criterion. In which case the inequalities are trivial. – Julien Apr 28 '13 at 03:17
  • I don't know of any reason to think the alternating series test applies in any nice way. I'm not completely sure the strict bounds come prepackaged this way for arbitrary order, either. For very small order, yes. – Will Jagy Apr 28 '13 at 03:32
  • @WillJagy Interesting. This seemed to be vonbrand's argument, that's why I thought it could be true. Well, this was instructive, thanks again. – Julien Apr 28 '13 at 03:35
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In case you do not understand vonbrand and user17762's reasoning, both are using the fact that $\int_1^n \frac{dk}{k}=\ln{n}$, and that a sum can be approximated by an integral (for large n values) with a correcting factor. Here the correcting factor is $\gamma$, the Euler-Mascheroni constant.

Zen
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You can look up "digamma function" for more info. A version I made up myself, which uses a symmetry property to give every other power, is $$ H_n = \log \left( n + \frac{1}{2} \right) + \gamma + \frac{1}{6(2n+1)^2} - \frac{7}{60(2n+1)^4} + \frac{31}{126(2n+1)^6} - \frac{127}{120(2n+1)^8} + \frac{511}{66(2n+1)^{10}} - O \left( \frac{1}{(2n+1)^{12}} \right) $$ with $$ \gamma = 0.5772156649015328606065... $$

I put part of this in my programmable calculator, just the 1/6 blah - 7/60 blahblah. Because of the alternating signs, this is definitive, too small with $n=4549,$ big enough with $n=4550.$ There is no guarantee the $\pm$ signs continue to alternate, and the rational coefficients are allowed to grow, note $511/66 \approx 7.74. $ The use of asymptotic expansions is to truncate at a convenient place and know enough about the possible error. Interesting to put this on a high precision program, with extremely high precision demanded for $\gamma,$ and look at the error of what I wrote multiplied by $(2n+1)^{12}.$

Will Jagy
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