Sum of $\sum\limits_{x=-\infty}^{\infty}x^{\operatorname{sign}(x)}$ Both the sum of $1+2+3+4+\cdots$ and the sum of $\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\cdots$ diverge. If both are paired together in one function, as seen above, can they amount to a number or can it even be calculated whatsoever?
 A: That would diverge. Just take $(1-\frac{1}{2})+(2-\frac{1}{3})+(3-\frac{1}{4})+\cdots>\frac{1}{2}+\frac{1}{2}+\cdots$, which also diverges.
A: It depends on what do you mean by: $$\sum\limits_{n = -\infty}^{\infty} n^{sign(n)}$$
Most people would say that:
$$\sum\limits_{n = -\infty}^{\infty} a_n=\sum\limits_{n = 0}^{-\infty}a_n+\sum\limits_{n = 1}^{\infty} a_n $$
where both converge. In this sense, it does not converge. If you want to take some type of principal value, it still doesn't converge. If you want to take a "principal value": 
$$\sum\limits_{n = -\infty}^{\infty} n^{sign(n)}=\sum\limits_{n=1}^{\infty} n-\frac{1}{n}$$ Which doesn't converge either. Try taking the limits different ways. It shouldn't be hard to convince yourself that it won't converge.
A: Let $f:\mathbb{N}\to\mathbb{N}$ be a function with $\lim_{n\to\infty}f(n)=\infty$.  One can define
$$S_n(f)=\sum_{k=-f(n)}^n k^{\text{sign}(k)}=-\left({1\over f(n)}+\cdots+{1\over2}+{1\over1}\right)+1+2+\cdots+n$$
and then ask for $\lim_{n\to\infty}S_n(f)$.  It's fairly clear you can tailor $f$ to get any limit you want.  In particular, $f(n)=\lfloor e^{n(n+1)/2}\rfloor$ should lead to the Euler-Mascheroni constant, or something like it.
