# 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 $N\to \infty$

• How do you define $\sum \limits_{n=-\infty}^{+\infty}a_n$? – Wojowu Jan 14 '16 at 17:28
• All the same terms are on both sides so the only issues can be definition and convergence. Suppose $a_0=0$ and $a_n=\frac{1}{n}$ otherwise: the left-hand side is then the sum of a countable number of $0$s but what is the right-hand side? – Henry Jan 14 '16 at 17:30
• Considering $a_n=n$ should demonstrate that the series needn't be the same. – MPW Jan 14 '16 at 17:30
• @Wojowu, 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 $N\to \infty$. – ZFR Jan 14 '16 at 17:31
• One often requires $$\lim_{M\to\infty,N\to\infty} \sum_{n=-M}^{N}a_n$$ to exist, which is stronger than only using symmetric limits. – MPW Jan 14 '16 at 17:35

## 3 Answers

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.)

• Thanks a lot! Nice answer! +1 – ZFR Jan 14 '16 at 17:52
• This is mathematically correct, of course, but in my experience taking symmetric limits is not the definition of convergence of a doubly-infinite sum, just as convergence of an integral over the entire line is not defined by convergence of the principal value. – Andrew D. Hwang Jan 14 '16 at 18:15
• @AndrewD.Hwang : Yes, I agree. +1. I was going to comment that the symmetric case is something of a P-value for the sum, and not what I would consider sufficient for convergence; however, that's what OP specifies. – MPW Jan 14 '16 at 19:37
• Ah right, (+1). :) I mis-read the OP as starting with the doubly-infinite series. – Andrew D. Hwang Jan 15 '16 at 2:55

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).

\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}

• This only works if we assume absolute convergence. – Wojowu Jan 14 '16 at 17:32
• You are changing terms in infinite series. I guess that it's illegal. – ZFR Jan 14 '16 at 17:32
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