Summation of 1/k [duplicate]

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 ?

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

$$\sum_{k = 1}^{n}{1 \over k} = \sum_{k = 0}^{n - 1}{1 \over k + 1} =\Psi\left(n + 1\right) - \Psi\left(1\right) = \gamma + \Psi\left(n + 1\right)$$ $\Psi\left(z\right)$ and $\gamma$ are the Digamma Function and the Euler-Mascheroni constant , respectively.
• @GerryMyerson In terms of the Gamma Function $\Gamma(z)$: $\Psi(z) = {\rm d}\ln\Gamma(z)/{\rm d}z$. – Felix Marin Mar 3 '14 at 8:18
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.
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.