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I saw this equation on a blackboard today: $\displaystyle \sum_{n=0}^\infty (-1)^n$

It got me thinking -- this must oscillate between $1$ and $0$, yes? So then does this sum even have a meaningful value?

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yes, the sequence of partial sum is not convergent. It does'nt converge. – matgaio May 16 '12 at 0:12
What you say you saw on the blackboard is not an equation. If you want a broad generic term, you could write "I saw this expression on the blackboard today:". Or of course you could just say "I saw this sum on the blackboard today:". – Michael Hardy May 16 '12 at 1:19
Related:… – Argon May 16 '12 at 2:17

This sum is called Grandi's series and does not converge in the usual sense. By usual sense, we mean that if you look at the sequence of partial sums then we get that $$S_n = \sum_{k=0}^{n}(-1)^k = \begin{cases} 1 & \text{if $n$ is even},\\ 0 & \text{if $n$ is odd}.\end{cases}$$ Hence, $\displaystyle \lim_{n \rightarrow \infty} S_n$ does not exist.

That said, convergence of partial sums is by no means the only way to define convergence. Another popular way of defining convergence is to look at the Cesàro sum. The Cesàro sum of the Grandi's series is $1/2$. The Cesàro sum is defined as the limit of the average of the sequence of partial sums i.e. if we let $$\tilde{S}_n = \dfrac{\displaystyle\sum_{m=0}^{n} S_m}{n+1},$$ then we have that $$\tilde{S}_n = \begin{cases} \frac12 & \text{if $n$ is odd},\\ \frac12 + \frac1{2n+2}& \text{ if $n$ is even},\end{cases}$$ which converges to $\dfrac12$.

There are also other ways to interpret the value of $\dfrac12$.

For instance, if we were to randomly choose a natural number and if we assign the probability of getting an odd number to be $\dfrac12$ and the probability of getting an even number to be $\dfrac12$, then the expected value of the sum $S_n$ is $\dfrac12$.

Another interpretation is through analytic continuation of the function $$f(x) = 1-x+x^2 - x^3 + \cdots, \, \lvert x \rvert < 1.$$ The function converges to $\dfrac1{1+x}$ on the interval $(-1,1)$. Not surprisingly, the analytic continuation of the function $f(x)$ on the entire complex plane is given by $f_{\text{ac}}(x) = \dfrac1{1+x}$, where $x \in \mathbb{C}\backslash \{-1\}$. Hence, if we plugin the value of $x=1$ in $f_{\text{ac}}(x)$, we get the value of $\frac12$. Hence, $$1 - 1 + 1 -1 + \cdots \underset{\text{ac}}{=} \frac1{2}.$$

There are also other regularization techniques like Borel summation, Ramanujan summation which assign a finite value to sums which do not converge. All these different techniques assign a value of $\dfrac12$ for the Grandi's series.

You can read more about Grandi's series here.


As @anon points out, the $\zeta$ regularization technique (which is actually closely related to Ramanujan summation technique) can also be used here to get the value of $1-1+1-1+\cdots$. Consider $$g(s) = 1 - \frac1{2^s} + \frac1{3^s} - \frac1{4^s} + \cdots$$ and $$f(s) = 1 + \frac1{2^s} + \frac1{3^s} + \frac1{4^s} + \cdots$$ The series $g(s)$ converges for Real$(s) > 0$ and converges absolutely for Real$(s) >1$. The series $f(s)$ converges for Real$(s) > 1$. In the region, Real$(s)>1$, we have that $$g(s) = 1 - \frac1{2^s} + \frac1{3^s} - \frac1{4^s} + \cdots = \left(1 + \frac1{2^s} + \frac1{3^s} + \frac1{4^s} + \cdots \right) - \left(\frac2{2^s} + \frac2{4^s} + \frac2{6^s} + \frac2{8^s} + \cdots \right) = f(s) - \frac1{2^{s-1}} f(s) = \left(1 - 2^{1-s} \right)f(s).$$

Now analytic continuation of $g(s)$ gives us $\eta(s)$ such that $\eta(s) = (1-2^{1-s}) \zeta(s)$, where $\zeta(s)$ is the analytic continuation of $f(s)$. Hence, plugging in $s=0$, we get that $$1 - 1 + 1 - 1 +\cdots \underset{\text{ac}}{=} \eta(0) = (1-2) \zeta(0) = -\zeta(0) = \frac12,$$ since the value of $\zeta(0) = -\dfrac12$.

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I didn't even know there was another notion of convergence in wich this little monster could converge. Learning something everyday =) – matgaio May 16 '12 at 0:22
Very thourough answer Marvis. – Pedro Tamaroff May 16 '12 at 0:54
It can also be regularized with the Dirichlet eta function $$\eta(s)=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^s}=(1-2^{1-s})\zeta(s)$$ analytically continued to $s=0$. – anon May 16 '12 at 1:11
when you write "\end{cases}." with the period outside of the "cases" environment, it looks like this: $\displaystyle\begin{cases} 1 & \text{if $n$ is even} \\ 0 & \text{if $n$ is odd}\end{cases}.$ ${}\qquad{}$ I changed it to $\displaystyle\begin{cases} 1 & \text{if $n$ is even}, \\ 0 & \text{if $n$ is odd}.\end{cases}$ ${}\qquad{}$ – Michael Hardy May 16 '12 at 1:23
@MichaelHardy Thanks for the edit. I shall follow this in future. – user17762 May 16 '12 at 1:26

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