# Conditional sum and absolute sum

My professor in a Measure theory class mentioned that there is a difference between $$\sum_{n\in\mathbb{N}}a_n$$ and $$\sum_{n=0}^\infty a_n,$$ in the sense that the latter considers the order (taking the definition of the bottom to be the usual limit of partial sums), while the former doesn't (defined as $\sup\{\sum_{a_n\in J}a_n : J \text{ finite} \}$).

I believe they are called conditional sums and absolute sums (but I could be mistaken).

I'm having trouble seeing the difference (i.e. I see the difference in the definiton, but I don't see how the two values could be different, or how a series $(a_n)$ could converge in one way, but not the other). Examples would be appreciated, please.

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See, e.g., here. – David Mitra Sep 12 '12 at 15:22

## 1 Answer

You've got the definitions right. The difference is simply that between absolute convergence and conditional convergence.

If all terms in the sum are nonnegative, then there is no difference between the values of these two sums: either both of them are the same finite number or both are $\infty$.

If $\sum_{n\in\mathbb{N}} |a_n|<\infty$ then $$\lim_{N\to\infty}\sum_{n=1}^N a_n = \sup\left\{\sum_{n\in I} a_n\,:\,I\text{ finite} \right\}$$ provided all terms are $\ge 0$.

If some terms are positive and some are negative and the sum of the absolute values is finite, then $\sum_{n\in\mathbb{N}} a_n$ is defined by treating the positive and negative terms separately, and then adding those two sums.

It is only when the sum of the positive terms and the sum of the negative terms are both infinite that there can be a difference. In that case the limit of finite sums may be a finite number, but it then becomes a different finite number if the order in which the terms are added is altered. The most well known example is the alternating harmonic series $$\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n} =\log_e 2.$$ If you rearrange the terms, you can make the limit of finite sums a different number. For example, if you add enough positive terms, the sum will exceed $10$, then add one negative term and the sum is less than $10$, then add more positive terms until the sum is more than $10$, then the next negative term, etc., and then the limit of partial sums is $10$.

Later note: Another example is that $$\int_0^\infty \frac{\sin x}{x}\,dx$$ exists as an improper integral, i.e. a limit as the upper bound of integration approaches $\infty$, but does not exist as a Lebesgue integral because the integrals of the positive and negative parts both diverge to $\infty$.

An easier example is $$\lim_{a\to\infty} \int_{-a}^a \frac{x\,dx}{1+x^2}\text{ versus }\lim_{a\to\infty} \int_{-a}^{2a} \frac{x\,dx}{1+x^2}.$$ The values of these two expressions can be found by freshman calculus methods, and they're different from each other.

Another, somewhat more involved, example that yields to freshman calculus methods is $$\int_0^1 \int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} \, dx \, dy \text{ versus } \int_0^1 \int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} \, dy \, dx.$$ Work them out and get two different numbers: $\pi/2$ and $-\pi/2$ (I don't recall which is which). Conclusion: The integrals of the positive and negative parts of this function must both diverge to infinity. This is an example of the difference between interated integrals and double integrals. These are iterated integrals. The double integral is an integral with respect to Lebesgue measure in the plane; the iterated integral is an iteration---one integral inside another---of two or more integrals with respect to Lebesgue measure on the line.

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Fascinating. Thanks. Quick question, in the case of that alternating harmonic series, what would be the value of $\sum_{n\in\mathbb{N}} a_n$? How would one go about calculating that? – FPP Sep 12 '12 at 15:48
No such value is defined when the positive and negative parts both diverge to infinity. Letting $f_+(x)=\left.\begin{cases} f(x) & \text{if }f(x)\ge 0, \\ 0 & \text{if }f(x)<0\end{cases}\right\}$, and $f_-(x)=\left.\begin{cases} -f(x) & \text{if }f(x)\le 0, \\ 0 & \text{if }f(x)>0\end{cases}\right\}$, so that $f=f_+ - f_-$, the Lebesgue integral $\int f$ is defined as $\int f_+ - \int f_-$, and is undefined when this takes the form $\infty-\infty$. – Michael Hardy Sep 12 '12 at 16:02