Summing infinitely many numbers: how to assign a value?

Question

If we take $S = 1-1+1-1+1-1+1-1+...$ we can show (in many different ways) that the result of the sum is $\frac{1}{2}$.

• One way for example would be to add $S$ to itself but shift it along one place,
  $2S =$
  $1-1+1-1+1-1+...$
  $0+1-1+1-1+1-...$

  Clearly if we add the two 'rows' together the sum of the first 'column' is 1 and the rest if the 'columns' sum to 0 resulting in:

  $2S = 1 \rightarrow S = \frac{1}{2}$

• Another would be to say that the sum, $S$, results in one if you stop at an 'odd' position in the sum i.e. $S = 1-1+1$ or $S=1-1+1-1+1$ we get $S=1$ whereas if you stop at an 'even' position i.e. $S = 1-1+1-1$ we get $S= 0$.

  So if we go till infinity (or arbitrarily far along the sum) the average answer will be the average of the two $S$ values, namely $\frac{1}{2}$

So now for my question(s):

• How does this really work, at what point does infinity 'take over' from the normal maths stuff.

• Clearly there are no fractions on the $RHS$ of $S$ yet there is one in the answer.

I am struggling to accept the answer, because our small albeit amazing brains think on a rather small scale relative to infinity.

Thanks a lot.

Answer 1

Remind that series like

$\displaystyle\sum_{k=0}^{\infty}t_k$

really aren't sums, but infinite sequences $(S_0,S_1,...)$, where

$S_n=\displaystyle\sum_{k=0}^{n}t_k$.

If the sequence $(S_0,S_1,...)$ is convergent, $\displaystyle\underset{n\rightarrow\infty}{lim}S_n$ exists and is denoted

$\displaystyle\underset{n\rightarrow\infty}{lim}S_n=\displaystyle\sum_{k=0}^{\infty}t_k$

The idea of limit is consistent and intuitive, but it is possible to dedicate other numbers to series (sequences) that isn't even convergent. The limit operator can be thought of as a function from convergent sequences to real numbers and in the same way one can define other functions from other classes of sequences to real numbers, as long as the functions are well defined.

Your examples has very little with sums to do, but functions like that seems to be of certain interest in science.

Answer 2

In the usual sense, $\sum_{n=0}^\infty (-1)^n$ is defined as $\lim_{m \to \infty} \sum_{n=0}^m (-1)^n$. This limit does not exist, so the infinite sum doesn't exist.

There are a number of "regularization" techniques for divergent infinite series, which can be used to assign meaning in a certain sense to such series. They do not, however, create results which cooperate nicely with the rules of arithmetic.

Perhaps the most popular technique is analytic continuation. In this case we have a series which really does converge for certain values of a complex parameter $s$. For example, $\sum_{n=1}^\infty n^{-s}$ converges in the usual sense provided $\text{Re}(s) > 1$. It is also analytic in the parameter $s$ on the same set. This set is open, and it is a theorem of complex analysis that an analytic function on a domain $\Omega$ is determined uniquely by its values on any open subset of $\Omega$. Therefore it is natural to ask what the maximal domain of an analytic function which agrees with the sum is, and to study the extension. The extension is called the Riemann zeta function, denoted $\zeta(s)$, and it is defined for all complex numbers except $1$. In particular we have $\zeta(-1)=-1/12$, which very informally means $1+2+3+\dots=-1/12$, surprisingly.

Another popular technique which is more straightforwardly applicable to the example in the OP is Cesaro summation. This basically amounts to adding up partial averages: we replace $\sum_{n=1}^\infty a_n$ with $\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \sum_{j=1}^k a_j$. An equivalent expression is $\lim_{n \to \infty} \sum_{j=1}^n \frac{(n-j+1)}{n} a_j$, which may clarify what is meant by partial averages.

If the original sum converges, then the Cesaro sum will be the same, but sums like the one in the OP may fail to converge in the usual sense but still converge in the Cesaro sense. Indeed note that $\sum_{j=1}^k (-1)^{j+1} = \frac{(-1)^{k-1} + 1}{2}$. Therefore $\sum_{k=1}^n \sum_{j=1}^k (-1)^{j+1} = \lceil \frac{n}{2} \rceil$, so the relevant limit is $\lim_{n \to \infty} \frac{1}{n} \lceil \frac{n}{2} \rceil$. The partial Cesaro sums are always between $\frac{1-1/n}{2}$ and $\frac{1+1/n}{2}$ so the limit is $\frac{1}{2}$.

Answer 3

My answer to your questions: it doesn't work, in the conventional sense.

I am well aware that there are ways to "show" that $1-1+1-1+1-1+\dots=\frac12$. In fact, in this answer, I show that $\eta(0)=\frac12$, where $\eta$ is the alternating $\zeta$ function $$ \eta(s)=\frac1{1^s}-\frac1{2^s}+\frac1{3^s}-\frac1{4^s}+\dots\tag{1} $$ Plugging in $s=0$ gives the series in question.

We can also note that for $x=1$, $\frac1{1+x}=\frac12$. We can then plug $x=1$ into the Taylor series $$ \frac1{1+x}=1-x+x^2-x^3+\dots\tag{2} $$ and get the series in question.

However, in the conventional sense, $(1)$ does not converge when $s=0$ and $(2)$ does not converge when $x=1$ because their terms do not tend to $0$. One problem with series that don't converge in the conventional sense is that they usually fail to obey some rule of arithmetic. For example, associativity. Addition is associative, which means that we can rearrange parentheses in any way we want and still get the same answer. Consider your sum grouped as follows $$ (1-1)+(1-1)+(1-1)+\dots\tag{3} $$ that is $$ 0+0+0+\dots=0\tag{4} $$ However, if we group the terms as follows $$ 1+(-1+1)+(-1+1)+(-1+1)+\dots\tag{5} $$ that is $$ 1+0+0+0+\dots=1\tag{6} $$ Thus, associativity fails. If you don't have associativity (and commutativity), then you can't add $$ 1-1+1-1+1-1+\dots\tag{7} $$ and $$ 0+1-1+1-1+1-\dots\tag{8} $$ to get $$ 1+0+0+0+0+0+\dots\tag{9} $$