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Is ${1\over 1}+{1\over 2}+{1\over 3}+\cdots +{1\over n}$ computable? I couldn't find any formulas to find the result. The explanation would be very helpful. Thanks before.

Martin Sleziak
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Jake Timberwood
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    please remove this queston, look up harmonic series before people starting to downvote for the harmonic series or harminc number for the n'th time – jimjim May 28 '13 at 10:19
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    I wonder who and why did downvote all in this question, which is not about the harmonic series but about the $;n$-th partial sum. Both answers btw look pretty nice to me. – DonAntonio May 28 '13 at 10:21
  • Why do you want a formula? There are useful asymptotic formulas: http://en.wikipedia.org/wiki/Harmonic_number#Calculation. – lhf May 28 '13 at 10:22
  • @lhf yeah i think it would be useful for the other questions related with this question – Jake Timberwood May 28 '13 at 10:25
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    @DonAntonio : Briefly it was me, but then I reverted them.\ – jimjim May 28 '13 at 10:39
  • Note that if $n$ is large, then the value is (by Euler-Macluarin) very close to $\int_1^n \frac{1}{x}dx + \text{const.} = \log(n) + \text{const.}$ – dohmatob Sep 21 '16 at 09:49

2 Answers2

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I think what you are looking for is the harmonic number.

Henrik Finsberg
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Yes. The result is: $$S=\Psi(n+1)+\gamma$$ where:$$S=\sum_{k=1}^n \frac{1}{k}$$ For the explanation see: http://en.wikipedia.org/wiki/Harmonic_number

Riccardo.Alestra
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