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This question already has an answer here:

I know that the infinite summation $$1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{n}+...$$ is divergent and also the sequence $$1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{n}-\ln n$$ is convergent.

I wonder that whether there exist a formula (closed form) for the obtain value of the following finite summation. $$1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{n}$$ If not why?

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marked as duplicate by J. J., mathlove, Najib Idrissi, apnorton, user147263 Aug 28 '14 at 11:57

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    $\begingroup$ It's not at all clear how this series is defined, unless the fourth term should be $\frac{1}{4}$, in which case no such closed form is known. $\endgroup$ – J. J. Aug 28 '14 at 11:08
  • $\begingroup$ in addition to @J.J. comment: check out the Harmonic series $\endgroup$ – Alex Aug 28 '14 at 11:09
  • $\begingroup$ @J.J. Thanks. I know that $\sum_{k=1}^n \dfrac{1}{k}> \ln (n+1)$ and $\sum_{k=1}^{2^n} \dfrac{1}{k}> 1+ \dfrac{n}{2}.$ I wont to know that why there does not exist a such closed form. Have any one prove that there can not be a such closed form? $\endgroup$ – Bumblebee Aug 28 '14 at 11:17
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    $\begingroup$ @Nilan I think you are looking for the $n^{th}$ harmonic number, $H_n$ but in the future, it would be better for you to explicitly state your series. $\endgroup$ – Jam Aug 28 '14 at 11:17
  • $\begingroup$ @Nilan the sequence $\sum_{k=1}^{\infty} (1/k-ln(k))$ is divergent, too-not convergent. $\endgroup$ – callculus Aug 28 '14 at 11:29
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$$H_n~=~\psi(n+1)+\gamma~=~-\int_0^1\ln\Big(1-\sqrt[n]x\Big)~dx~=~\int_0^1\frac{1-x^n}{1-x~~}~dx,$$ where $\psi$ is the digamma function, and $\gamma$ is the Euler-Mascheroni constant. See harmonic numbers for more details.

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    $\begingroup$ OP $\displaystyle 1+\frac12+\frac13+\cdots+\frac1n=\sum_{k=1}^n\frac1k=H_n$ $\endgroup$ – Ali Caglayan Aug 28 '14 at 11:49
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$$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots+\frac{1}{n}=\sum_{i=1}^{n} \frac{1}{i}=H_n$$

where $H_n$ is the $n^{th}$ harmonic number.

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