Sum of the Harmonic Series? What happens when you use summability methods on the harmonic series? 
I'm quite surprised I haven't been able to find anything on this anywhere, considering that the partial sums of the harmonic series grow at a logarithmic rate, while series whose partial sums grow quadratically are summable. 
 A: The Ramanujan summation of the reciprocal of the positive integers is equal to the Euler-Mascheroni constant. Scroll down to the bottom of the page until where it says
$$\sum_{n\geq1}^\Re \frac{1}{n}=\gamma$$
A: 
What happens when you use summability methods on the harmonic series? 

Most of them (Cesaro, Euler, Noerlund, Borel) diverge. I don't know whether the Aitken process can be made to give something.
The condition which allows the Cesaro/Euler/Noerlund/Borel to assign a value to a divergent series is when it has terms with alternating signs, or complex numbers. If they are applied to divergent series with strictly positive terms they all go to infinity. One must try whether there are possibilities for functional relations between nonalternating and alternating series which can then allow to sum the alternating series instead of the nonalternating one and then to recalculate the result using that functional relation (as it is done with the geometric series via the rational expression as a fraction ${ 1 \over 1-q}$ (except the pole at $q=1$) or with the Zeta-series and the functional relation with the alternating Zeta series in the way L. Euler had introduced it and is later made by the methods of analytic continuation).
A: Harmonic series is essentially $\zeta(1)$, Riemann Zeta function evaluated at $1$. While the function has a pole at $1$, we can find its Cauchy principal value there:
$$\lim_{h\to0}\frac{\zeta(1+h)+\zeta(1-h)}2=\gamma$$
It turns out to be Euler-Mascheroni constant.
A: and how about generalized harmonic series ??
there are two possible results
$$ \sum_{n=0}^{\infty} \frac{1}{(n+a)} = -\Psi (a) $$
and $$ \sum_{n=0}^{\infty} \frac{1}{(n+a)} = -\Psi (a) +log(a) $$
if $ a=1 $ we recover the euler mascheroni constant
