calculus - On the equivalence of two definitions for summations over the integers

If $a_k\in\mathbb{C}$, then define

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

Question: Is it true that

$$\sum_{k=-\infty}^{\infty}a_k=L\\ \Updownarrow\\ \sum_{k=0}^{\infty}a_k\text{ and }\sum_{k=1}^{\infty}a_{-k}\text{ both exist and }\sum_{k=0}^{\infty}a_k+\sum_{k=1}^{\infty}a_{-k}=L$$

Thoughts:$[\Downarrow]$ Couldn't come up with something useful. Tried to show that both series are Cauchy. Failed.

$[\Uparrow]$ Let

$$\alpha:=\sum_{k=0}^{\infty}a_k\\ \beta:=\sum_{k=1}^{\infty}a_{-k}$$
Given $\epsilon>0$, there exist $N_1,N_2$ such that
$$\left|\sum_{k=0}^na_k-\alpha\right|<\frac{\epsilon}{2}\\ \left|\sum_{k=1}^ma_{-k}-\beta\right|<\frac{\epsilon}{2}$$
whenever $n>N_1$ and $m>N_2$. Hence if $N:=\max\{N_1,N_2\}$ then
$$\begin{align} \left|\sum_{k=-m}^na_k-L\right|&=\left|\sum_{k=0}^na_k-\alpha+\sum_{k=1}^ma_{-k}-\beta\right|\\ &\leq\left|\sum_{k=0}^na_k-\alpha\right|+\left|\sum_{k=1}^ma_{-k}-\beta\right|\\ &<\frac{\epsilon}{2}+\frac{\epsilon}{2}\\ &=\epsilon \end{align}$$
whenever $m,n>N$.

Answer

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.