# Is there any summation method that assigns $\sum_{n=1}^\infty \frac{1}{n} =-\frac{\pi}{2}$

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

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

• Standard regularizations give $'\zeta(1)'=\gamma$, the Euler Mascheroni constant. – Olivier Oloa Apr 28 '16 at 21:43
• 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 $\zeta (4)$ – Alfred Yerger Apr 28 '16 at 21:49
• @OlivierOloa : ok, show us how with a 'standard regularization method' you obtain $\gamma$ – reuns Apr 28 '16 at 21:53
• Have alook here (Ramanujan summation):mathoverflow.net/questions/64898/… and en.wikipedia.org/wiki/Ramanujan_summation – Olivier Oloa Apr 28 '16 at 22:16
• @AlfredYerger Lol, the unsolved mysteries of math, eh? – Simply Beautiful Art Apr 28 '16 at 22:25

$$\zeta(1)=\sum_{n\ge1}^{\Re}\frac1n=\lim_{N\to\infty}\left(\sum_{n=1}^N\frac1n-\int_1^N\frac1tdt\right)=\gamma$$
And no, I do not think you can assign $-\frac\pi2$ to the series.
• this is not $\zeta(1)$ but $\lim_{s \to 1} \zeta(s) - \frac{1}{s-1}$, the derivation is on wikipedia :) – reuns Apr 29 '16 at 0:11