# Is it possible to use regularization methods on the Harmonic Series?

I recently learned about summation methods when dealing with divergent series to give them a finite value. An example of this isusing Cesàro summation on Grandi's series to get 1/2. However every method I know of is unable to sum the harmonic series. Are there any summation methods that work on the harmonic series or is it provably impossible to sum this series?

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What does it mean when one says "it is provably impossible to sum a series"? Interestingly, the Riemann-Zeta function $\zeta(z)$ has a pole at $z=1$. – Aryabhata Feb 2 '11 at 7:14
(Sorry I meant this as a comment, never used this site before, if someone can fix this that would be great) What I meant by "is it provably impossible" is if it was still an open question or if someone had shown that it couldn't be done by any method. Honestly, since I'm new to this I don't even know if its possible to show it can't be done. The fact that the Riemann-Zeta function has a pole at 1 is one of the reasons I asked this question, if Zeta function regularization can solve 1 + 2 + 3 + 4 ... as -1/12. Why can't it sum the harmonic series? – user6522 Feb 2 '11 at 11:35
I asked this question on MO awhile ago: mathoverflow.net/questions/3204/… – Qiaochu Yuan Feb 2 '11 at 11:35

I'll stick to real, positive exponents $s$ in $\sum_{n=1}^\infty n^{-s}$. The following regularization works for $s\in (0,\infty)\setminus \{1\}$: formally, $$(1-2^{1-s})\sum_{n=1}^\infty n^{-s} = \sum_{n=1}^\infty n^{-s} - 2\sum_{n=1}^\infty (2n)^{-s} = \sum_{n=1}^\infty (-1)^{n-1}n^{-s} \tag1$$ where the right-hand side of (1) converges (by the alternating series test) for $s>0$. Thus, for $s\in (0,\infty)\setminus \{1\}$ we can interpret $\sum_{n=1}^\infty n^{-s}$ as $$\zeta (s)=(1-2^{1-s})^{-1}\sum_{n=1}^\infty (-1)^{n-1}n^{-s} \tag2$$ Although setting $s=1$ in (2) does not work, we can do simple averaging: $$\sum_{n=1}^\infty n^{-1} \,=" \lim_{\epsilon\to 0} \frac{\zeta(1-\epsilon)+\zeta(1+\epsilon)}{2} =\gamma \tag3$$ where $\gamma$ is the Euler-Mascheroni constant, $\gamma=0.57721\dots$ Indeed, the averaging in (3) cancels out the contribution of the simple pole in the Laurent series $$\zeta(s)=\frac{1}{s-1}+\gamma+O(s-1)$$
you can use the regulator $R(n,s)= \frac{n^{s}+n^{-s}}{2}$ so the series $\sum_{n=1}^{\infty}R(n,s)n^{-1}$ converges to the Euler-Mascheroni constant in the limit $s \to 0$ – Jose Garcia Jul 4 '13 at 10:01