Assigning values to divergent series

Question

I have been looking at divergent series on wikipedia and other sources and it seems people give finite "values" to specific ones. I understand that these values sometimes reflect the algebraic properties of the series in question, but do not actually represent what the series converges to, which is infinity. Why is it usefull to assign values to divergent series?

The only theory I could come up with, is this:

Say you have 2 divergent series, series' A and B, and you assign each a value,

Series ($A= \sum_{n=0}^\infty a_n$), which I assigned the value Q

and series ($B= \sum_{n=0}^\infty b_n$ ), which I assigned the value P

But it just so happens that series $C=A-B= \sum_{n=0}^\infty (a_n-b_n)$ converges. Could that imply that the actual value of series $C$ is the difference of the two assigned values to $A$ and $B$, that is $\sum_{n=0}^\infty (a_n-b_n)=Q-P$ ?

If so, then that would make some sense to me, as to why people sometimes assign values to divergent series.

Answer 1

The most common situation with a divergent series is this: an infinite series with a variable $z$ is given, which converges for some values of $z$ in the complex plane $\mathbb C.$ On the region of convergence, the series defines a holomorphic function, call it $f(z).$ Then the analytic continuation of the function $f(z)$ is correctly described. As a result, there is a well-defined value $f(z)$ for $z$ values that would cause the original series to diverge.

The best example is ZETA. I guess Euler found values of $\zeta(-n)$ at negative integers, and wrote these down in the style of divergent series. So people get an impression that one assigns a value to a divergent series by clever manipulation. This is not the general case, however. When the radius of convergence is strictly exceeded, we are simply reporting the value given by the analytic continuation. Not assigning.

Here is an elementary example: Let us take $$ f(z) = \frac{1}{1 + z^2}. $$ Now, for $|z| < 1,$ we know $$ f(z) = 1 - z^2 + z^4 - z^6 + z^8 - z^{10} \cdots $$ If I wrote $$ 1 - 9 + 81 - 729 + 6561 - 59049 \cdots = \frac{1}{10} $$ you would have every reason to be suspicious as the series obviously diverges. But if I instead wrote $$ f(3) = \frac{1}{10} $$ you would think that was probably alright.

Answer 2

There exist systems in which "divergent" series "coverge". They are called the systems of $p$-adic numbers, and for example in $3$-adic system you have

$$\sum_{k\geq0} 2\cdot 3^k=-1,$$

therefore

$$-1=\dotsm 2222222\bullet_3.$$

To show that the identity above makes some sense, let's try to multiply $(-1)\cdot(-1)$ by the "basic shool" algorithm, in the base $3$. We get

$$\begin{array} {}&{}&\dotsm&2&2&2&2&2&2&2\\ {}&\times&\dotsm&2&2&2&2&2&2&2\\\hline {}&{}&\dotsm&2&2&2&2&2&2&1\\ {}&+&\dotsm&2&2&2&2&2&1\\ {}&+&\dotsm&2&2&2&2&1\\ {}&+&\dotsm&2&2&2&1\\ {}&+&\dotsm&2&2&1\\ {}&+&\dotsm&2&1\\ {}&+&\dotsm&1\\ {}&+&\vdots&\vdots\\ \hline {}&=&\dotsm&0&0&0&0&0&0&1\\ \text{where the carry was:}&{}&\dotsm&6&5&4&3&2&1&0\\ \end{array}$$

Answer 3

I understand that these values sometimes reflect the algebraic properties of the series in question, but do not actually represent what the series converges to, which is infinity.

The problem here is about the choice we have in defining the value of infinite series. The definition of addition does not provide us with a definition for this, because this will only fix the definition of the sum of a finite number of terms. With the rules of addition that tell you how to add up two numbers and the associative property, you can add up 3 numbers, 4 numbers etc. and this way you can only ever add up finite number of terms.

This means that we need to come up with a definition of the sum of infinite series. The standard choice of defining this to be the limit of the partial sums only defines the sum if this limit exists, i.e. if the series converges. If this limit does not exit, i.e. if the series is divergent, then nothing is defined. If we want to define a quantity A and say that A = B and it turns out that B isn't defined, then we didn't actually define A at all!

It's then not a contradiction if by some other means we do manage to define the sum of a divergent series. It doesn't contradict with the partial sums tending to infinity. Given a series, one can e.g. interpret the value of that series to be given by the value taken by a function whose series expansion generates the given series. This amounts to performing an analytic continuation using that function, as explained in the answer given by @WillJagy.

If we stick to this interpretation, then we should note that the relevant theorems used when performing an expansion like e.g. Taylor's theorem do not tell you to sum the series using the standard definition of the limit of the partial sums, they instead tell you that the value of the function you're computing is given by a finite number of terms of the series plus a remainder term.

In general, we then don't know that the remainder term is. In case the series converges, the limit of the remainder term must then exist. This limit should then be zero if the function that upon expansion generates the series is analytic. In case the series is divergent, one can invoke an argument based on analytic continuation to get a handle on the remainder term. Doing so leads to a generalized formula for the value of a series. As explained here, if the partial sums are $S(N)$ then the standard definition of the sum $S$ of convergent series:

$$S = \lim_{n\to\infty} S(N)$$

generalizes to:

$$S = \operatorname*{con}_N\int_{N-1}^N S(x)dx$$

where $\displaystyle \operatorname*{con}_N $ denotes extraditing the constant term in the large $N$ expansion of the expression.

The interpretation of the value of the series is then that it gives you the value a function whose series expansion generates the series, where we then n assume that this function is maximally analytic in the sense that any nonanalytic behavior is dictated by the series.

If the difference of two divergent series is a covergent series and the values of the divergent series are defined using analytic continuation from covergent series, then the identity theorem implies that difference of the values of the divergent series will equal the value of the convergent series.

Answer 4

The regularized values of divergent series are not always what we should think of as the values of those sums. When a series diverges to infinity, the regularized value is only the finite part of the series.

One can map divergent series to surreal numbers, and the regularized sul will be the finite part.

There is quite developed theory of mapping Hardy fields to surreals, and basically considering surreals as an H-field (H-field is a Hardy field with a unity).

In simple language, Hardy fields are the fields of germs (asymptotic behaviors) of functions at infinity. For instance, the germ of the function $f(x)=x$ is greater than any real numbers but smaller than the germs of $x+1$, $2x$ or $x^2$.

The surreal numbers also can be seen as a Hardy field, with canonical embedding of the germs of smooth functions into surreals being equalizing $\omega$ with the germ of the identity function $f(x)=x$.

This way, you can embeed germs at infinity into surreals, and with some other limitations, divergent improper integrals (at least those which correspond to surreals with no infinitesimal part).

With series the matter is a bit more complicated, since you need to convert the discrete partial sum into a smooth germ. But this formula (which is similar to the one in the answer by Saibal Mitra) does the trick:

$$\sum_{k=0}^\infty a_k=\int_{\omega }^{\omega +1} \left(\sum _{k=0}^{x-1} a_k\right) \, dx$$

This way you can get the expression of a divergent series as a surreal number, compare them, extract the finite part (that is regularized value), etc.

For instance, you can see that $\sum_{k=0}^\infty e^k=e^{\omega }+\frac{1}{1-e}$.

You can see that $e^\omega$ is a log-atomic number, so it has zero finite part, so the second term is the regularized value.

In a similar way, you can see that $\sum_{k=0}^\infty k=\frac{\omega ^2}{2}-\frac{1}{12}$ and so on.

For the sum of harmonic series you will get $\sum_{k=1}^\infty \frac1k=\ln \omega+\gamma$. For some reason, online Wolfram Alpha is too weak for this, but Wolfram Mathematica does just well.