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I have the inequation : $\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}>2012$ and I need to show that there is a n which satisfies it. How do I do it? I tried to show that it is bigger then $\frac{1}{1}+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2^n}$ which is equal to $1-\frac{2^n-1}{2^{n-1}}$. Then I wanted to show that there is an n which makes that sum be bigger than 2012. Is it correct? I found somewhere that it says : $1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^m}>1+\frac{m}{2}$ and then it says that $n=2^{4024}$. Why it is like that?

Bram28
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Ghost
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2 Answers2

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Just show the sequence diverges. That means there is an $n$, even if you can't point to one.

It diverges because you can keep finding groups of terms that add up to 1/2 or more:

1 > 1/2

1/2 = 1/2

1/3+1/4 > 1/4 + 1/4 = 1/2

1/5+1/6+1/7+1/8 > 1/8+1/8+1/8+1/8 = 1/2

(then take the next 8 terms .. then the next 16 ... etc)

So the whole sum is greater than $ 1/2+1/2+1/2+1/2+ ... = \infty$

And with this method, you can point to an $n$ for which it is $> 2012$: we need 4024 of these groupings (since we need 4024*1/2=2012), and $m$ groupings take $2^{m-1}$ terms:

The first grouping (1) took 1 term,

the next (1/2) also 1 term,

the next (1/3+1/4) 2 terms,

the next 4 terms, etc., ...

so the first 3 groupings require 4 terms, for 4 groupings we need 8 terms, for 5 we need 16, and now you see the pattern: we need $2^{m-1}$ terms to get $m$ groupings of terms where each grouping sums up to 1/2 or more.

So for $n=2^{4023}$ we know it has surpassed 2012.

Bram28
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  • How do I do that? – Ghost Mar 17 '17 at 20:12
  • The typical way to do that with this one is to point out that we can keep finding groups of terms that add up to 1/2 or more ... that is 1 > 1/2, 1/2 = 1/2, 1/3 + 1/4 > 1/2, 1/5+1/6+ ... > 1/2 (do you see how far you have to go? and how you can keep finding those groupings?) – Bram28 Mar 17 '17 at 20:15
  • @OvyOvy the most common "simple" proof is that $\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\dots+\frac{1}{8}+\frac{1}{9}+\dots+\frac{1}{16}+\dots $ $> \frac{1}{1}+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{8}+\dots+\frac{1}{8}+\frac{1}{16}+\dots+\frac{1}{16}+\dots$ $=1+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\dots$ – JMoravitz Mar 17 '17 at 20:17
  • Note that the approach will give an $n$ in the order of $2^{4000}$, while the true bound is between $2^{2901}$ and $2^{2902}$. – Arthur Mar 17 '17 at 20:25
  • How do you find how many $\frac{1}{2}$ are depending on n? – Ghost Mar 17 '17 at 20:32
  • @OvyOvy Just added some more explanation to my answer – Bram28 Mar 17 '17 at 20:33
  • I do not get it. I guess it's too hard for me. Thanks anyway. – Ghost Mar 17 '17 at 20:37
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    Ok. After I read abit and wrote it down I managed to get it. Thank you. – Ghost Mar 17 '17 at 20:43
  • @OvyOvy Yay! Glad I could help ... I am sure there are better explanations than mine ... but working it out yourself is always best! – Bram28 Mar 17 '17 at 20:45
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You are looking for $n$ such that $$S=\sum_{i=1}^n\frac 1i=H_n>2012$$ FOr sure, $n$ will be large; so, considering the asymptotics of the harmonic numbers $$H_n=\gamma +\log \left({n}\right)+\frac{1}{2 n}+O\left(\frac{1}{n^2}\right)$$ and ignoring higher order terms, you need to solve for $n$ $$\gamma +\log \left({n}\right)+\frac{1}{2 n}=2012$$ the solution of which being given in terms of Lambert function $$n=-\frac{1}{2 W\left(-\frac{1}{2} e^{\gamma -2012}\right)}$$ But, for small $x$, $W(x)\approx x$ which makes $$n\approx e^{2012-\gamma}\approx 355\times 10^{871}$$ and for this value $S_n=2012.00095$.

Computing tables $$\left( \begin{array}{cc} n & H_n \\ 350 \times 10^{871} & 2011.98676 \\ 351 \times 10^{871} & 2011.98962 \\ 352 \times 10^{871} & 2011.99246 \\ 353 \times 10^{871} & 2011.99530 \\ 354 \times 10^{871}& 2011.99813 \\ 355 \times 10^{871} & 2012.00095 \\ 356 \times 10^{871} & 2012.00376 \\ 357 \times 10^{871} & 2012.00657 \\ 358 \times 10^{871} & 2012.00936 \\ 359 \times 10^{871} & 2012.01215 \\ 360 \times 10^{873} & 2012.01494 \end{array} \right)$$

Claude Leibovici
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