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I'm working on a problem in probability and got to the sum $\sum_{k=0}^{{n-1}}\frac1{1-\frac kn}$, where $n$ is constant.

I tried changing its form but didn't get anywhere.

Any hint?

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  • $\begingroup$ I think you made a mistake in the index of summation. $\endgroup$ – hamid kamali Dec 12 '15 at 23:33
  • $\begingroup$ Multiply the fraction by $n/n$ and look up "harmonic number". $\endgroup$ – user147263 Dec 12 '15 at 23:34
  • $\begingroup$ @hamidkamali what do you mean? $\endgroup$ – Whyka Dec 12 '15 at 23:36
  • $\begingroup$ @NormalHuman oh, that seems interesting. Thank you! $\endgroup$ – Whyka Dec 12 '15 at 23:36
  • $\begingroup$ consider $\frac{1}{1-\frac{k}{n}}$ when $k=n$. $\endgroup$ – hamid kamali Dec 12 '15 at 23:38
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$$\sum_{k=0}^{n-1}\frac1{1-\frac k n}=\sum_{k=0}^{n-1}\frac n{n-k}=\sum_{m=1}^{n}\frac n{m}=n\sum_{m=1}^{n}\frac 1{m}=n\cdot H_n$$

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  • $\begingroup$ I just did this, but wondering how to calculate this $H_n$. Using wikipedia for the meanwhile... Thanks anyway $\endgroup$ – Whyka Dec 12 '15 at 23:45
  • $\begingroup$ As @Normal Human suggested. en.wikipedia.org/wiki/Harmonic_number $\endgroup$ – Kay K. Dec 12 '15 at 23:45
  • $\begingroup$ There is no closed form. But it's "roughly" $\ln n$, when $n\to\infty$. $\endgroup$ – Clement C. Dec 12 '15 at 23:46
  • $\begingroup$ Yes, I did this after reading what he suggested $\endgroup$ – Whyka Dec 12 '15 at 23:46
  • $\begingroup$ @ClementC. Thanks! This is what I was looking for $\endgroup$ – Whyka Dec 12 '15 at 23:46

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