I have the following question from Function Theory of One Complex Variable - Greene/Krantz:

Give an example of a series of complex coefficients $ a_n$ such that $\lim_{N \to + \infty} \sum_{n= -N}^{N} a_n$ exists but $\sum_{-\infty}^{+\infty} a_n$ does not converge.

The answer key I have says that $a_n = n$ answers the question.

I understand that $\lim_{N \to + \infty} \sum_{n= -N}^{N} n = \lim_{N\to+\infty}[-N + (-N+1) +...+ -1 +0+1+...(N-1)+N] = 0$, as can be seen from each term cancelling.

However, on the question of why $\sum_{-\infty}^{+\infty} n$ does not converge I'm a little stumped. Thinking about it intuitively, wouldn't you expect the same sort of cancellation of terms?

If anyone can provide a formal proof of why this series doesn't converge, I'd be very grateful. Thanks in advance!

  • $\begingroup$ what's the definition of "$\sum_{-\infty}^{+\infty} a_n$ converges" ? $\endgroup$ – mercio May 5 '16 at 10:31
  • 1
    $\begingroup$ @mercio The definition is as follows (Greene/Krantz p. 110): For the doubly infinite series $\sum_{j=-\infty}^{\infty} a_j$, we say that such a series converges if $\sum_{j=0}^{\infty} a_j$ and $\sum_{j=1}^{\infty} a_{-j}$ converge in the usual sense... which means you have indeed answered my question :) Thanks! $\endgroup$ – K.Reeves May 5 '16 at 10:39

This is similar to the difference between integral and Cauchy principal value

For example you know that $$\int _ {\mathbb R} x dx$$ does not exists, but

$$\lim_{R \to \infty} \int_{-R}^R x dx = 0$$

which is the Cauchy principal value.

The main point is that $\int_\mathbb R$ is not defined as a limit $\lim_{R \to \infty} \int_{-R}^R$.. For the riemann integral it is defined as

$$\lim_{a \to -\infty} \int_a^0 x dx + \lim_{b \to +\infty} \int_{0}^b x dx$$

and you can already see that neither of those converge.

A similar difference is there with the Lebesgue integral; The integral of a function $f(x)$ is defined as

$$\int f(x) dx = \int f(x) ^+ dx - \int f(x)^-dx$$ where $f(x)^+ = \max(f(x), 0)$ and $f(x)^- =- \min(f(x), 0)$. The main point is that we want the two separate integral to exists, and only then (if both exists) we sum them up. In this way there can't be any cancellation like the one happening with Cauchy principal value.

Your case is clearly in the same spirit, although it depends on how you define the double series $\sum_{-\infty}^{\infty}$. This is either defined as a double limit (in the "spirit" of the riemann integral) or as an integral with respect to the counting measure on $\mathbb Z$ (so basically it's a Lebesgue integral)


This isn't a formal proof, but rather an intuitive reason for it not being absolutely convergent. If you start adding at 0 then add one term at a time on each side, you get

$$s_0 = 0$$ $$s_1 = -1 + 0 + 1 = 0$$ $$s_2 = -2 + -1 + 0 + 1 + 2 = 0$$ $$\cdots$$ $$s_k = 0$$

but there's no reason to start summing at $0$. Instead, if you start summing at 1 and do the same process, you get

$$t_0 = 1$$ $$t_1 = 0 + 1 + 2 = 3$$ $$t_2 = -1 + 0 + 1 + 2 + 3 = 5$$ $$\cdots$$ $$t_k = 2k+1$$

and we have unbounded partial sums! In fact, there's no reason we have to sum one term from each side for each partial sum. We could be summing 1 from the right, 1 from the left, 2 from the right, 1 from the left, 3 from the right, 1 from the left and so on so that we still sum every $n$ but our partial sums now grow superlinearly.


I'm reading from the same book right now actually! Greene and Krantz's definition for the convergence of a doubly infinite series like this is that $\sum_{-\infty}^{+\infty}\alpha_j$ converges if both of the series $\sum_{0}^\infty\alpha_j$ and $\sum_{1}^\infty\alpha_{-j}$ converge.

To see why $\{j\}_{-\infty}^{+\infty}$ doesn't converge with this definition, we just notice that the series $\sum_0^\infty j$ doesn't converge and neither does $\sum_1^\infty -j$. Of course the limit $\lim_{N\to\infty}\sum_{-N}^Nj$ exists and equals $0$.


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