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There is a partial sum formula for $$\sum_{x=1}^n x^1 = \frac{n(n+1)}{2}$$ and even one when the exponent of $x$ is $0$: $$\sum_{x=1}^n x^0 = n$$ but I cannot find one for exponent $-1$: $$\sum_{x=1}^n x^{-1} = ?$$

I tried $$\frac2{n(n+1)},$$ but that failed miserably.

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No, there is no nice closed form for the harmonic numbers. There are some very accurate approximations that are easily computed;

$$H_n\approx\ln n+\gamma+\frac1{2n}-\frac1{12n^2}$$

is quite good, where $\gamma\approx 0.5772156649$ is the Euler-Mascheroni constant.

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  • $\begingroup$ Could you provide a source for that approximation please? $\endgroup$
    – zerosofthezeta
    Sep 17 '13 at 6:35
  • $\begingroup$ @zerosofthezeta: It’s at the link, at the bottom of the Calculation section of the article. $\endgroup$
    – Brian M. Scott
    Sep 17 '13 at 6:36
  • $\begingroup$ Got it, thanks! $\endgroup$
    – zerosofthezeta
    Sep 17 '13 at 6:37
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    $\begingroup$ I've found a better approximation (for my purposes) using: ln(x + 0.5) + 0.5772156649 $\endgroup$
    – trr
    Jun 1 '16 at 2:18
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    $\begingroup$ @H.R.: I wouldn't call either the integral representation or the alternating summation a closed form, but they do give alternative ways to look at $H_n$ that can be useful. $\endgroup$
    – Brian M. Scott
    Jul 13 '16 at 1:53

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