Does $a_0+\sum \limits_{n=1}^{\infty}(a_n+a_{-n})=\sum \limits_{n=-\infty}^{+\infty}a_n$? Let we have series $a_0+\sum \limits_{n=1}^{\infty}(a_n+a_{-n})$. Why it's equal to $\sum \limits_{n=-\infty}^{+\infty}a_n$? 
I guess that for this manipulation me must know something about convergence.
Can anyone explain it rigorously please?
P.S. this topic related with Fourier series when they define series $\sum \limits_{-\infty}^{+\infty}c_ne^{inx}$. So I guess that this summation understand as limit of $\sum \limits_{n=-N}^{N}a_n$ as $N\to \infty$
 A: In the case of requiring symmetric limits, they are the same. Just consider the partial sums:
$$S_N = a_0 + \sum_{n=1}^{N} (a_n + a_{-n})$$
and
$$T_N = \sum_{n=-N}^{N} a_n$$
both are the sum of exactly the same $N+1$ terms. So the limits are identical (or fail to exist identically). (Recall that an "infinite sum" isn't really a sum -- it is always a limit of partial sums.)
A: In general,
$$
\sum_{n=-\infty}^{\infty} a_{n}
  = \sum_{n=-\infty}^{-1} a_{n} + \sum_{n=0}^{\infty} a_{n}
  = \sum_{n=1}^{\infty} a_{-n} + \sum_{n=0}^{\infty} a_{n},
\tag{1}
$$
in the sense that the doubly-infinite series converges if and only if both singly-infinite series converge. When that is the case, the standard theorem about adding convegent series guarantees the sum is
$$
a_{0} + \sum_{n=1}^{\infty} (a_{-n} + a_{n}).
\tag{2}
$$
As multiple comments note, there are doubly-infinite sequences (such as $a_{-n} = -a_{n}$, with $(a_{n})$ your favorite non-summable sequence) for which (2) is not equal to to (1) (because (2) converges and (1) does not).
A: $$\begin{align}
a_0+\sum \limits_{n=1}^{\infty}(a_n+a_{-n}) &= a_0 + a_1 + a_{-1} + a_2 + a_{-2} + \ldots\\
&=a_0 + a_1 + a_2 + \ldots + a_{-1} + a_{-2} + \ldots\\
&=a_0 + \sum \limits_{n=-\infty}^{1} a_n +\sum \limits_{n=1}^{\infty} a_n\\
&=\sum \limits_{n=-\infty}^{+\infty} a_n\\
\end{align}$$
