Divergent Series

Consider these sums:

\begin{equation} \label{eq1}
\sum_{n=0}^\infty {(1)^n} = 1+1+1+1+1+ \cdots
\end{equation}

\begin{equation} \label{eq2}
\sum_{n=0}^\infty {(-1)^n} = 1-1+1-1+1-1+ \cdots
\end{equation}

\begin{equation} \label{eq3}
\sum_{n=0}^\infty {2^n} = 1+2+4+8+16+ \cdots
\end{equation}

Is there a way to give a meaning to these sums?  That is, to sum them?

What if there was?  Imagine that we have a machine $ \mathfrak{S}$ that can sum infinite numbers of numbers. What would such a thing give for these sums? Let's consider what basic sanity we might expect. To start, it seems like anything that adds should maintain this kind of consistency:

\begin{equation*}
\mathfrak{S} (a+b+c+d+ \cdots) = a+ \mathfrak{S} (b+c+d + \cdots) \tag{property 1}
\end{equation*}

This is sometimes called stability. In addition, let's require our machine can maintain linear behavior:

\begin{equation*}
\mathfrak{S} \left[ \sum_n ( \alpha  a_n + \beta  b_n) \right] = \alpha \, \mathfrak{S} \left( \sum_n a_n \right) + \beta \, \mathfrak{S} \left( \sum_n b_n \right)   \tag{property 2}
\end{equation*}

Let's apply our machine to \eqref{eq1}: $ s  = \mathfrak{S} (1+1+1+ \cdots) $

Using property 1, we can write $ s  = 1 + \mathfrak{S} (1+1+1+ \cdots) $. So then $ s = 1 + s $. Seems ridiculous, unless $ s $ is infinity, which is what we might expect for this sum. So maybe things are working out so far.

Let's look at \eqref{eq2}.

\[ s = \mathfrak{S}(1-1+1-1+1-1+\cdots) \]

\[ s = 1+ \mathfrak{S}(-1+1-1+1-1+\cdots) \]

\[ s = 1- \mathfrak{S}(1-1+1-1+1-1+\cdots) \]

So then $ s = 1-s $, or $ s=1/2 $.  Well that's disappointing.  The individual terms of the sum bounce around interminably, the partial sums do the same, but our $ \mathfrak{S}$ gives a small positive rational number.  This is getting weird.  Did we go wrong somewhere?  The consistency properties we assumed were pretty modest.

Let's look at \eqref{eq3}:
\[ \sum_{n=0}^\infty 2^n = 1 +2 +4  +8 +16 + \cdots \]

\[ s = 1 + \mathfrak{S}(2+4+8+16+ \cdots) \]

\[ s = 1 + 2  \mathfrak{S}(1+2+4+8+ \cdots) \]

So $ s= 1 + 2s $, and $s =-1$.

Wow.  The terms of this sum are, each, bigger than in \eqref{eq2}, but now we are getting a finite number and it's negative.  Let's try one more.

\[ s = \mathfrak{S} (1 -2 +3 -4 +5 -6 +7 -8 + \cdots ) \]

\[ s= 1+ \mathfrak{S}(-2+3-4+5-6+7-8 + \cdots) \]

Add these two together (we're using property 2 this time):

\[ 2s = 1 + \mathfrak{S}(-1+1-1+1-1+1-1+ \cdots) \]

\[ 2s = 1 - \mathfrak{S}(1-1+1-1+1-1+1 -1 +\cdots) \]

From our result for \eqref{eq2}, we can write $ 2s = 1 - \frac{1}{2}$, or $ s = \frac{1}{4} $.

All of these sums are divergent -- the partial sums do not converge.  Perhaps divergent sums are just trouble and we're better off not trying to find meaning in that demonic mess.  Let's back up to safer ground and examine an infinite sum that behaves nicely and converges.

This sum

\begin{equation*}
\sum_{n=1}^\infty {\frac{1}{n}} \tag{harmonic}
\end{equation*}

(famously) diverges.  But this one

\begin{equation*}
\sum_{n=1}^\infty {\frac{(-1)^{n-1}}{n}} \tag{alternating harmonic}
\end{equation*}

converges, very slowly, to $ ln(2)$. After a moment's consideration you may see that the alternating harmonic series contains two sums, in a sense.  One has an infinite number of positive terms, $ 1 + \frac{1}{3} + \frac{1}{5} +\cdots$ and the other has an infinite number of negative terms $ - \frac{1}{2} - \frac{1}{4} - \frac{1}{6} -\cdots$. So if we are allowed to associate terms at will, I can make the sum converge to any number I want.  Say, 50.  I just sum terms from the positive set until I get just over 50, then start using terms from the negative set till I get under 50, then use positive terms...

So it should be clear that we can't insist on associativity when working with infinite sums.  That is, addition is not infinitely associative. This is surprising. If readers were wondering why we didn't allow the infinite summing machine $ \mathfrak{S}$ to be associative, now you know.

Okay then, it seems unavoidable that infinite sums are weird, and convergence doesn't relieve the issue.  Something spooky is going on.  Can we come up with a concrete example of something that behaves like $ \mathfrak{S} $?  The following approach, due to Euler, might illuminate things.

Euler Summation

Consider an infinite sum $ s = \sum_n a_n$, and form a function $ f(x) $ like so:

\[  f(x) = \sum_n a_n x^n \]

The Euler sum is

\[ \lim_{x \to 1-}f(x) \]

Let's try this on \eqref{eq2}.

\[ Euler(1-1+1-1+1-1+\cdots) = 1-x + x^2-x^3 + \cdots = \lim_{x \to 1-}\frac{1}{1+x} = \frac{1}{2} \]

Well that's nice, it's consistent with our previous approach. But maybe we're just lucky.  Let's try something else.

Borel Summation

Consider $ \int_0^\infty \frac{t^n}{e^t}dt = n!$, so that

\[ 1= \int_0^\infty \frac{t^n}{{e^t}{n!}}dt\]

Form a new series
\[ \sum_n a_n {\int_0^\infty \frac{t^n}{{e^t}{n!}}dt}\]

The Borel sum is what you get if you reverse the order of summation and integration:

\[ \int_0^\infty dt \, e^{-t} \sum_n \frac{t^n a_n}{n!}\]

Now look at $ Borel(1-1+1-1+1-1 + \cdots)$.

\[ \int_0^\infty dt \, e^{-t} \sum_n \frac{{(-t)}^n}{n!} \]

\[ \int_0^\infty e^{-2t} dt = \frac{1}{2}\]

Same!

General Summation

Returning to our general summation approach, let's investigate the impact of this slight modification to \eqref{eq2}:
\[ s = \mathfrak{S} (1 +0 -1 +1 +0 -1 +1 +0 -1 + \cdots) \]
\[ s = 1+ \mathfrak{S} (0 -1 +1 +0 -1 +1 +0 -1 + \cdots) \]
\[ s = 1+ \mathfrak{S} (-1 +1 +0 -1 +1 +0 -1 + \cdots) \]
Adding these three relations together: $ 3s=2 +  \mathfrak{S}(0+0+0+0+0+\cdots) $, or $ s=2/3 $.  So inserting zeros into the sum changes the result from what we got for \eqref{eq2}.  It's hard to rely on intuition in this area.  In fact, we might now question what really constitutes the "basic sanity" we started with.

Zeta Function Regularization

Consider the Zeta function:  $ \zeta(z) = \sum_{n=1}^\infty \frac{1}{n^z}$.  Zeta is remarkable for many reasons and the first is that it is a meromorphic function with one and only one simple pole, at $ z=1 $.  At that point, $\zeta(1)$ is the harmonic series we met earlier, which we were not able to assign a sum to. But look what $\zeta$ says about $z=0$:

\[\zeta(0) = \sum_{n=1}^\infty \frac{1}{1} = 1+1+1+1+1+1+1+ \cdots \]

$ \zeta(z)$ is analytic at $z=0$, and in fact $\zeta(0) = - \frac{1}{2} $.  What does this mean, especially after the happy result we got from general summation?  Well, $\zeta(z)$ must be either not stable or not linear, or both.  Whatever it is, that a perfectly analytic function assigns a value to this sum is not easy to dismiss.

There is a whole field of study that goes under the name summation theory that explores summation strategies and axioms.