Regularize the sum: $\sum_{n=0}^\infty (-1)^{n+1}({2+4n})$

$$\sum_{n=0}^\infty (-1)^{n+1}({2+4n})=2-6+10-14\cdots$$ is a clearly divergent sum, which I am trying to regularize. I need to figure out this sum because I have been trying to solve the sum integral $$\sum_{n=0}^\infty\int_{n^2}^{(n+1)^2} \sin \pi\sqrt x\text{ d}x=\color\red{\sum_{n=0}^\infty \frac{(-1)^{n+1}({2+4x})}{\pi}}$$ Thanks for any help.

I can't really figure it out because it can not be represented in terms of the Riemann Eta $\eta$ function.

• typo $x\to n$ perhaps? – Pierpaolo Vivo Jan 23 '16 at 19:51
• @PierpaoloVivo Nice find, but only of the first equation. – user266519 Jan 23 '16 at 19:53
• and the title as well – Pierpaolo Vivo Jan 23 '16 at 19:54
• @PierpaoloVivo Lol. And that too. – user266519 Jan 23 '16 at 19:54
• Sinisa Bubonja has done some interesting work in furthering these sort of sums (the work describing the proof and actual method is in Serbian though, so be warned... The results are the best I've seen though) link – Brevan Ellefsen Jan 24 '16 at 0:24

$$\sum_{n=0}^\infty (-1)^{n+1}({2+4n})=0$$
$$\sum_{n=0}^\infty (-1)^{n+1}({2})=-1/2*2=-1$$ $$\sum_{n=0}^\infty (-1)^{n+1}({4n})=1/4*4=1$$ $$\sum_{n=0}^\infty (-1)^{n+1}({2+4n})=0$$
$$\sum_{n=1}^\infty (-1)^{n+1}({2-4n})=0$$ $$\sum_{n=1}^\infty (-1)^{n+1}({2})=1/2*2=1$$ $$\sum_{n=1}^\infty (-1)^{n+1}({4n})=-1/4*4=-1$$