5
$\begingroup$

The notation $\sum_{k=1}^\infty a_k$ always means $$\lim_{n\rightarrow\infty}\sum_{k=1}^n a_k.$$

What about $\sum_{k=-\infty}^\infty a_k$, such as in the Laurent series? Does it always means $$\lim_{n\rightarrow\infty}\sum_{k=-n}^n a_k,$$ or does the meaning depend on the context?

$\endgroup$
9
$\begingroup$

The notation must be used only when $\dagger$ is satisfied. $$\lim_{N_1 \to \infty} \lim_{N_2 \to \infty} \sum_{k=-N_1}^{k=N_2} a_k \text{ and }\lim_{N_2 \to \infty} \lim_{N_1 \to \infty} \sum_{k=-N_1}^{k=N_2} a_k \text{ exists and }$$ $$\lim_{N_1 \to \infty} \lim_{N_2 \to \infty} \sum_{k=-N_1}^{k=N_2} a_k = \lim_{N_2 \to \infty} \lim_{N_1 \to \infty} \sum_{k=-N_1}^{k=N_2} a_k \tag{$\dagger$}$$ If $\dagger$ is not satisfied, the notation doesn't make sense. Sometimes, it can also be used if both $$\lim_{N_1 \to \infty} \lim_{N_2 \to \infty} \sum_{k=-N_1}^{k=N_2} a_k \text{ and }\lim_{N_2 \to \infty} \lim_{N_1 \to \infty} \sum_{k=-N_1}^{k=N_2} a_k$$ are either $+ \infty$ or $- \infty$.

$\endgroup$
6
+50
$\begingroup$

The notation $$ \sum_{k=-\infty}^{\infty} a_k$$ has the potential to be ambiguous, and if you use it, you should give sufficient context describing what you mean by it. In particular, you need to know what it means for such a series to converge. Lacking such context, I think that the most reasonable interpretation is that such a series converges if and only if $$ \sum_{k=0}^{\infty} a_k \qquad\text{and}\qquad \sum_{k=1}^{\infty} a_{-k} $$ converge absolutely. If both series converge absolutely (in the usual sense), then we say that $$ \sum_{k=-\infty}^{\infty} a_k = \lim_{M\to\infty} \sum_{k=0}^{M} a_k + \lim_{N\to\infty} \sum_{k=1}^{N} a_{-k}. \tag{1}$$ This is typically how Laurent series are understood; both the positive and negative series must converge individually in order to say that the series converges as a whole—this is the notion of convergence which is given by the Wikipedia article cited in the question (and also the definition given by MathWorld), and is a stronger condition than requiring the existence of the limit $$ \lim_{n\to\infty} \sum_{k=-n}^{n} a_k. $$

More generally, you might drop the condition that both series in (1) converge absolutely, and simply require that they converge. You lose some properties which you might like to have, but there are gains, too. You could also allow for infinite limits, but you have to be careful here—if both series in (1) diverge to $+\infty$ or both series diverge to $-\infty$, then we might say that the two-sided series diverges to $\pm \infty$. However, if one diverges to $+\infty$ and the other to $-\infty$, the two-sided series doesn't make much sense. One possible convention is to require that all of the terms be nonnegative, then allow for infinite series.

Note that (1) is very similar to the definition of the value of an improper Riemann integral of the form $$ \int_{-\infty}^{\infty} f(x)\,\mathrm{d}x. $$ The value of this integral is given by $$ \int_{-\infty}^{\infty} f(x)\,\mathrm{d}x = \lim_{a\to\infty} \int_{0}^{a} f(x)\,\mathrm{d}x + \lim_{b\to-\infty} \int_{b}^{0} f(x)\,\mathrm{d}x. $$ (Note that we can replace $0$ in the integrals above with any constant $c$. Similarly, we can replace $0$ and $1$ in the lower limits of the series in (1) with $L$ and $L+1$ for any integer $L$.)

$\endgroup$
  • $\begingroup$ +1. I think the most valuable element in this answer is the comparison with the improper Riemann integral. $\endgroup$ – Ewan Delanoy Jul 13 '19 at 7:23
  • $\begingroup$ This seems compatible with @user17762's answer. Is it? $\endgroup$ – Olli Niemitalo Jul 13 '19 at 19:00
  • $\begingroup$ @OlliNiemitalo Yes. It is the same idea (though I am insisting on absolute convergence, which is a little different). $\endgroup$ – Xander Henderson Jul 13 '19 at 20:25

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.