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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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    I asked this question on MO awhile ago: http://mathoverflow.net/questions/3204/does-any-method-of-summing-divergent-series-work-on-the-harmonic-series2011-02-02

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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 $\zeta(s)=\frac{1}{s-1}+\gamma+O(s-1)$

(Adapted from MathOverflow).

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    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 $2013-07-04