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$$$ \sum_{k = 1}^{n}{1 \over k} = \sum_{k = 0}^{n - 1}{1 \over k + 1} =\Psi\left(n + 1\right) - \Psi\left(1\right) = \gamma + \Psi\left(n + 1\right) $$ $\Psi\left(z\right)$ and $\gamma$ are theDigamma Function and the Euler-Mascheroni constant , respectively.
$\endgroup$ 1 $\begingroup$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.
$\endgroup$ $\begingroup$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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