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What is summation of 1/k where n ranges from 1 to n. I need the general formula for the summation. I know the series tends to infinity when k tends to infinity . But upto n terms there must be a definite sum value. Correct ?

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    $\begingroup$ It has a definite value, but there is no closed-form formula for that definite value in terms of standard elementary functions. $\endgroup$ – Gerry Myerson Mar 3 '14 at 8:06
  • $\begingroup$ are you sure.. ? Somewhere I remember seeing it as gamma + ln(n) where gamma is euler-mascheroni constant $\endgroup$ – Adeetya Mar 3 '14 at 8:07
  • $\begingroup$ That's an approximation, true in the limit. See also math.stackexchange.com/questions/37496/… $\endgroup$ – Gerry Myerson Mar 3 '14 at 8:09
  • $\begingroup$ @Adeetya The formula you are thinking of is $H_n=\gamma+\psi_0(n+1)$ $\endgroup$ – David H Mar 3 '14 at 8:09
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$$ \sum_{k = 1}^{n}{1 \over k} = \sum_{k = 0}^{n - 1}{1 \over k + 1} =\Psi\left(n + 1\right) - \Psi\left(1\right) = \gamma + \Psi\left(n + 1\right) $$ $\Psi\left(z\right)$ and $\gamma$ are the Digamma Function and the Euler-Mascheroni constant , respectively.

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  • $\begingroup$ @GerryMyerson In terms of the Gamma Function $\Gamma(z)$: $\Psi(z) = {\rm d}\ln\Gamma(z)/{\rm d}z$. $\endgroup$ – Felix Marin Mar 3 '14 at 8:18
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This partial sum defines the $n$-th "harmonic number". You can find lots of information on these numbers by looking up that term. But no simple and exact closed formula.

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Moreover if you want to compute with some easy steps the $n$-th Harmonic Number, there is a demonstration that asserts that $H_n = \sum_{k = 1}^{n}{1 \over k} = \ln(n) + \frac{1}{2n} + \gamma$, where $\gamma$ is, as said before, the Euler-Mascheroni constant.

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  • $\begingroup$ You do not need to use TeX to get something like italics, you can use markdown. $\endgroup$ – Martin Sleziak Mar 3 '14 at 18:33

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