# Assigning values to divergent series

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.

• One short answer is that a series that diverges in one sense can converge in another sense. For example, the average of the first $n$ partial sums may converge as $n\to\infty$, as in the case of $\sum_{n=0}^\infty (-1)^n$. IIRC, when Poincaré found the power-series solutions of differential equations arising in astronomy, they had radius of convergence zero, but they were useful. – Michael Hardy Dec 1 '12 at 20:54
• Whats a radius of convergence, also is my theory wrong? – Ethan Dec 1 '12 at 21:01
• A power series $\sum_{n=0}^\infty a_n z^n$ converges if $|z|$ is smaller than a number that depends on the sequence $\{a_n\}_{n=0}^\infty$, and diverges if $|z|>R$. That number $R$ is the radius of convergence. If $R=0$, then the series diverges unless $z=0$ (in which case the series is $a_0+0+0+\cdots$, which obviously converges). (In some cases $R=\infty$, so the series converges no matter how big $z$ is.) – Michael Hardy Dec 1 '12 at 21:27

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.

• @OldJohn, thank you, kind sir. Barcelona was a blowout. Real-Atletico is pretty close but ill-tempered. No red cards yet but not for lack of trying. – Will Jagy Dec 1 '12 at 22:21
• +1 for the excellent explanation - I've read 2 'popular mathematics' books about the Riemann Zeta function and neither was particularly clear on this point. – user3490 Dec 1 '12 at 23:02
• @OldJohn, It is easier for a camel to go through the eye of a needle, than for gravy to pass through a tea strainer. – Will Jagy Dec 1 '12 at 23:57

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

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

• I wonder why the downvote. – yo' May 1 '15 at 19:07