# A divergent series i can't manage to sum

I know that this series is not summable in the usual sense so I'm "playing the game" of divergent series summation. The series is:

$$\sum_{n=0}^{+\infty}-\frac 12(1+3(-1)^{n+1})=1-2+1-2+1-2+\dots$$

I tried Abel summation method which failed this cause:

$$\sum_{n=0}^{+\infty}-\frac 12(1+3(-1)^{n+1})x^n=1-2x+x^2-2x^3+\dots=\frac {1-2x}{1-x^2}$$

Has singularities at $x=\pm 1$

Then i tried Borel summation but again I get to:

$$\sum_{n=0}^{+\infty}-\frac 12(1+3(-1)^{n+1})x^n=\frac 12(\frac 1{x-1}-\frac 3{x+1})$$

For $x\in ]-1;+\infty[-\{1\}$ that gives the same problem, and I don't think this is the case for Ramanujan summation.

I started to try Cesaro summation and I got to the following sequence of partial sums:

$$1;-1;0;-2;-1;-3;-2;-4;\dots$$

And the sequence of Cesaro means seems to diverge:

$$1;0;0;-\frac 12;-\frac 35;-1;-\frac 87;-\frac 32;\dots$$

I really don't know what other method should I try.

• The partial sums do not alternate around some value but decrease infinitely with only a constant disturbance so "averaging" to any level of recursion of the partial sums shall not provide an upper (positive or negative) bound and thus you cannot expect, that Cesaro, Euler or Borel-sum can give a finite value. – Gottfried Helms Mar 1 '16 at 8:08
• The sum of -1 you know. And the sum of (-1)^n you know. Now multiply the first one by an half. And the second one by 3. Add them together and solved. If not, The sum of -1 you need to find an equality to an alternating series, Dirichlet eta function seems a popular pick. And the sum of (-1)^n is solveable by the methodes you mentioned. I myself would just start at n=1 instead of n=0. – Gerben Jan 13 '17 at 2:35

$$\sum_{n=1}^\infty -\frac{1}{2} (1 + 3 (-1)^n) n^{-s} =\zeta(s) - \dfrac{3}{2^s} \zeta(s)$$ and taking $s=0$ gives us the value $1$.
• I think one can think of this series as $\zeta(0)-2\zeta(0)=-\zeta(0)=\frac 12$ but is this allowed ? – Renato Faraone Feb 29 '16 at 21:37