What is the sum of $$\frac12+ \frac13+\frac14+\frac15+\frac16 +\dots\frac{1}{2012}+\frac{1}{2013} $$
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This is essentially what is called a harmonic number, and there are no closed forms for it (although there are good approximations) – Mariano Suárez-Álvarez May 18 '13 at 07:44
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Your number is $1$ less than the Harmonic Number $H_{2013}$. The linked article explains how to approximate this using the logarithm, and gives better approximations. – André Nicolas May 18 '13 at 07:46
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A closed expression is $\displaystyle{\large H_{n} = \Psi\left(n + 1\right) + \gamma}$. See $\displaystyle{\large\bf\mbox{6.3.2}}$ in this table. – Felix Marin Jul 31 '14 at 00:42
3 Answers
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Not really an interesting question, I fear ...
The exact answer is $\frac{A}{B}$, where $A$ is the 873-digit number
$$\begin{align} &1457823020375738882085285521838680376351317016179578495942597370\\ &0378320415510098781465535340716404121275266248327296358362232678\\ &6278734102622894043296263865840984434831046940807020935913518007\\ &6610170430184975025171489501039855136340044946766172114345187347\\ &2525258101947297211093973306792764671036527339494431750637074683\\ &4230708772438305435131071598225161542024364751281758482910493532\\ &5272679486529134147964088721658417270988342669017294670746222437\\ &5396289282709392060556292768642859506675716675390362111468228302\\ &0782368060066857220428567636667351228472222194509213532517098318\\ &4215312313922564589890501941681764939250636021442764267189354097\\ &6967312781103861891868999835897977944251765509632397992463764707\\ &2757539640418390812471213879592790834646976169124993489059261606\\ &2602876187541263449739774491663332258026609790766133246441424484\\ &757471204138450579265132408964711222835023\\ \end{align}$$
and $B$ is the 872-digit number
$$\begin{align} &2029024882178588859070537995606394892063138617868032497494729352\\ &4603117588071109954985856100075037278586690682862226370908784076\\ &1373758182460385685570815580907584983469652245087586518224049939\\ &8538189200052467764823885503628456029563567766917663813400872806\\ &6581413837126629349013539786812278235204246553266267204666411453\\ &3233291465507116087957721742791662582248630820924043188884944781\\ &6407535614283438262641159101052916955096435138537168813836363531\\ &9456214006726533268868764071437503549723903396960519957683391782\\ &6358829094665518541761643350701837468138883360957730534259985433\\ &0461170299850901340867381633962516193737332035551187788058506648\\ &9862570396408121482924662574281092579439686439440266363727224229\\ &3000091507088457965475315993866302318289137980710843830365091268\\ &4178987345171765323904440193241153257507552225072663118382054707\\ &65887615406383472808692605376427780480000\\ \end{align}$$
Also, the first digits are 7.184845455 ...
achille hui
- 122,701
Ewan Delanoy
- 61,600
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5+1 for the stamina . You have a mistake in the 568-th digit of the second number...Just kidding (perhaps), but how would you prove me wrong? – DonAntonio May 18 '13 at 10:20
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it's easy to prove , just write a simple python code which will give you the same answer – Mostafa 36a2 Jan 11 '14 at 19:07
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2@Mostafa36a2 sorry, but this will just instigate flame wars about computer-assisted proofs. – PrimeRibeyeDeal Jul 31 '14 at 01:27
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For sufficiently large $N$:
$$\sum_{n=1}^N \frac{1}{n} = \log{N} + \gamma + \frac{1}{2 N}+O\left(\frac{1}{N^2}\right)$$
Plugin $N=2013$ and subtract 1. Note that $\gamma \approx 0.577216$. This approximation for this value of $N$ is good to about 3 significant figures.
Ron Gordon
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$$ H_{2013} - 1 = \left[\Psi\left(2014\right) + \gamma\right] - 1 = \Psi\left(2014\right) + \left(\gamma - 1\right) \approx {\tt 7.1848}\ $$
Felix Marin
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