Sum ideas (part 1)

[I]t is broadly true to say that mathematicians before Cauchy asked not “How shall we define 1 – 1 + 1 – …?” but “What is 1 – 1 + 1 – …?”, and that this habit of mind led them into unnecessary perplexities and controversies which were often really verbal.

— G. H. Hardy, Divergent Series

What is the value of an infinite series?

I mean, what does “value” mean here?

With a finite series, you can (in principle) just add all the numbers together. You take the result and call that the value of the series. The value of 1+2+3+4+5 is 15. No problem. But you can’t do that with an infinite series. You’d never complete the process — and so you can’t get its result.

You learn about infinite series in school, and what you learn is that some series converge to a limit. That is, if you have the infinite series

S = \displaystyle \sum_{k=0}^{\infty} a_k

it converges if, loosely speaking, the sequence of partial sums

S_n = \displaystyle \sum_{k=0}^{n} a_k

approaches some value s arbitrarily closely as n gets larger; that value is called the limit and we can write

\displaystyle \lim_{n\to\infty} \sum_{k=0}^{n} a_k = s.

And then there’s a bit of a leap as we regard s as not just the limit of the partial sums, but as the value of the infinite series:

\displaystyle \sum_{k=0}^{\infty} a_k = s

But not all series have limits, and if the limit doesn’t exist, then we regard the value of the infinite series as undefined.

That all can be learned in an intuitive way, though it can be made much more formal, and on the surface it makes sense. But there’s a sort of swindle going on here. You’re thinking of the “value” of the series as “what you would get if you added up all the infinite number of numbers”… but you can’t add them all up, so how can you assert what you would get if you did?

Here’s a demonstration of why that way of thinking about it is deceptive: the Riemann series theorem. Consider the series

\displaystyle 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - ...

You can show this converges to the limit \ln 2. But you can take the same series and rearrange the terms like this:

\displaystyle \left (1 - \frac{1}{2} - \frac{1}{4}\right ) + \left (\frac{1}{3} - \frac{1}{6} - \frac{1}{8}\right )  + \left (\frac{1}{5} - \frac{1}{10} - \frac{1}{12}\right )...

and that converges to \frac{1}{2}\ln 2. You’re adding up “the same” numbers in the second series as in the first, and addition is commutative, so you should get the same answer if you could add up all the numbers in the series (which you can’t)… but the limits are different! In fact, Riemann proved that any conditionally convergent series (one that has a limit, but the sum of the absolute values of the terms does not) can be rearranged to give you any limit, including \infty or -\infty, or no limit at all. It seems the “value” of a conditionally convergent series is a property not just of the numbers being summed, but the order they’re being summed in, and that’s not at all true of the value of a finite sum. Really the limit is a number that can be unambiguously associated with an infinite series, and we can define it as that series’s value, but “value” here means something different than in the case of a finite series where it’s just what you get if you add all the terms together. In fact, what the value means is just what it is: the limit of the sequence of the finite sums.

That being said, what about a divergent series?

Stay tuned.

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s