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I don't know too much about alternate summation methods, but am interesting to know if any give the sum of the harmonic series to be

$$-\frac{\pi}{2}$$

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    $\begingroup$ Standard regularizations give $'\zeta(1)'=\gamma$, the Euler Mascheroni constant. $\endgroup$ – Olivier Oloa Apr 28 '16 at 21:43
  • $\begingroup$ I'm not sure if this is your motivation or not, but I once thought about this, and the reason was because I wanted to know if there was some kind of factorial behavior going on in the denominator. The value of $\zeta(3)$ is unknown, but even if it worked, the pattern stops there, as it does not hold for $\zeta (4)$ $\endgroup$ – Alfred Yerger Apr 28 '16 at 21:49
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    $\begingroup$ @OlivierOloa : ok, show us how with a 'standard regularization method' you obtain $\gamma$ $\endgroup$ – reuns Apr 28 '16 at 21:53
  • $\begingroup$ Have alook here (Ramanujan summation):mathoverflow.net/questions/64898/… and en.wikipedia.org/wiki/Ramanujan_summation $\endgroup$ – Olivier Oloa Apr 28 '16 at 22:16
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    $\begingroup$ @AlfredYerger Lol, the unsolved mysteries of math, eh? $\endgroup$ – Simply Beautiful Art Apr 28 '16 at 22:25
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$$\zeta(1)=\sum_{n\ge1}^{\Re}\frac1n=\lim_{N\to\infty}\left(\sum_{n=1}^N\frac1n-\int_1^N\frac1tdt\right)=\gamma$$

It equals the Euler-Mascheroni constant by definition.

You can find the said formula and derivation on the wikipedia.

And no, I do not think you can assign $-\frac\pi2$ to the series.

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    $\begingroup$ this is not $\zeta(1)$ but $\lim_{s \to 1} \zeta(s) - \frac{1}{s-1}$, the derivation is on wikipedia :) $\endgroup$ – reuns Apr 29 '16 at 0:11
  • $\begingroup$ @user1952009 I think that this is (exactly) the spirit: removing the divergent part. $\endgroup$ – Olivier Oloa Apr 29 '16 at 5:23
  • $\begingroup$ @user1952009 Well, if you want, you can try it however you wish and post it as your answer. $\endgroup$ – Simply Beautiful Art Apr 29 '16 at 12:18

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