On the equivalence of two definitions for summations over the integers

Solution 1:

I assume you meant (adding quantors for $m,n$):

$$\forall\epsilon>0,\exists N,\forall m > N, \forall n> N\implies\left|\sum_{k=-m}^na_k-L\right|<\epsilon.$$

The [$\Uparrow$] case is OK. For [$\Downarrow$], let $N(\epsilon)$ denote a suitable $N$ for $\epsilon$ in the above condition. If $n_2 > n_1 > N(\frac{\epsilon}{2})$, then $$\left|\sum_{n=n_1+1}^{n_2} a_n\right| = \left|\sum_{n=-n_1}^{n_2} a_n - \sum_{n=-n_1}^{n_1} a_n\right| = \left|\left(\sum_{n=-n_1}^{n_2} a_n -L \right) - \left(\sum_{n=-n_1}^{n_1} a_n - L\right)\right| \le \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.$$

This means $\sum_{n=0}^{\infty} a_n$ is a Cauchy series. The proof for the negative part being Cauchy is done just the same.

Proving that $L$ is now the sum of the limits of the positive and negative parts works just as in the [$\Uparrow$] part.

Adding remarks due to comment from OP below: That argument seems to work as well, but I was literally looking at your prove for $[\Uparrow]$. You start there by defining $\alpha$ and $\beta$. In that $[\Uparrow]$ proof that they exist is the assumption, in the $[\Downarrow]$ proof this was proven by them being Cauchy series'. Then you choose $\epsilon$ and based on that $N_1,N_2$. Then you use $L$ for the one and only time, but it is simply used as $L= \alpha + \beta$. So you proved that the $L$ defined that way fits your definition for "sum of series from $-\infty$ to $\infty$". Since that value is 'obviously' unique if it exists at all, you have now proven (in the $[\Downarrow]$ proof) that the value $L$ (which we assumed existed) must be equal to $\alpha + \beta$, which what was left to show.