1
$\begingroup$

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 $N\to \infty$

$\endgroup$
  • 1
    $\begingroup$ How do you define $\sum \limits_{n=-\infty}^{+\infty}a_n$? $\endgroup$ – Wojowu Jan 14 '16 at 17:28
  • $\begingroup$ 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? $\endgroup$ – Henry Jan 14 '16 at 17:30
  • $\begingroup$ Considering $a_n=n$ should demonstrate that the series needn't be the same. $\endgroup$ – MPW Jan 14 '16 at 17:30
  • $\begingroup$ @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 $N\to \infty$. $\endgroup$ – ZFR Jan 14 '16 at 17:31
  • $\begingroup$ 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. $\endgroup$ – MPW Jan 14 '16 at 17:35
2
$\begingroup$

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.)

$\endgroup$
  • $\begingroup$ Thanks a lot! Nice answer! +1 $\endgroup$ – ZFR Jan 14 '16 at 17:52
  • 1
    $\begingroup$ 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. $\endgroup$ – Andrew D. Hwang Jan 14 '16 at 18:15
  • $\begingroup$ @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. $\endgroup$ – MPW Jan 14 '16 at 19:37
  • $\begingroup$ Ah right, (+1). :) I mis-read the OP as starting with the doubly-infinite series. $\endgroup$ – Andrew D. Hwang Jan 15 '16 at 2:55
2
$\begingroup$

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).

$\endgroup$
0
$\begingroup$

$$\begin{align} a_0+\sum \limits_{n=1}^{\infty}(a_n+a_{-n}) &= a_0 + a_1 + a_{-1} + a_2 + a_{-2} + \ldots\\ &=a_0 + a_1 + a_2 + \ldots + a_{-1} + a_{-2} + \ldots\\ &=a_0 + \sum \limits_{n=-\infty}^{1} a_n +\sum \limits_{n=1}^{\infty} a_n\\ &=\sum \limits_{n=-\infty}^{+\infty} a_n\\ \end{align}$$

$\endgroup$
  • 3
    $\begingroup$ This only works if we assume absolute convergence. $\endgroup$ – Wojowu Jan 14 '16 at 17:32
  • $\begingroup$ You are changing terms in infinite series. I guess that it's illegal. $\endgroup$ – ZFR Jan 14 '16 at 17:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.