# Grandi's Series; tends to $1/2$, but why is this considered a valid sum?

Grandi's series,

$$1+1-1+1-1+1-1+...$$

can be expressed as the below:

$$\sum_{n=0}^\infty(-1)^n$$

Two valid sums that make sense to me are $1$, and $0$, depending on how you approach the series. $(1+1)-(1+1)-(1+1)-...=0$, and $1+(1-1)+(1-1)+(1-1)+...=1$.

There is consensus, however, that the actual sum is $\frac{1}{2}$. Why? I understand the approach of finding partial means of the series, and they do indeed tend to $\frac{1}{2}$, but it seems unintuitive to assert that the sum is neither $1$ or $0$.

A more convincing method I found was assuming the series is $S$, then shifting it such that $S-1 = S$, then through algebra finding $S = \frac{1}{2}$, but again, it seems more intuitive answer is either $0$ or $1$. I say this strictly because adding and subtracting integers should equal an integer, never a fraction.

Is this a characteristic of infinite series, which is not specific to Grandi's series?

• This is a divergent series - so it does not have an "actual sum". The sum you refer to is known as the Cesàro sum which is an alternative means of assigning a sum to an infinite series which may (or may not) converge. Oct 29, 2013 at 22:38

Consider power series $$S(x) = \sum_{n=0}^{\infty} (-1)^n x^n, \qquad x\in [0;1).$$ It is geometric series: $$\sum_{n=0}^{\infty} (-x)^n = \frac{1}{1-(-x)} = \frac{1}{1+x}.$$

So, $$S(x)=\frac{1}{1+x}, \qquad x\in[0,1).$$

$S(x)$ is continuous and bounded on $[0;1)$. So, we can find limit: $$S = \lim_{x\to 1} S(x) = \frac{1}{2}.$$

See Abel summation for better understanding.

The series does not converge it is not mathematically valid if we look at the epsilon definition of convergence. The result is useful in physics and is used there.

Obviously the series actually doesn't converge to $$\frac{1}{2}$$. But it is oscillating between $$1$$ and $$0$$. But we can assign this a value $$\frac{1}{2}$$ because it is most reasonable. I want to give a very reasonable proof for this. The proof is done with Fourier series.

I can give you a very beautiful proof with the help of Fourier series. For a function like this, $$f(t)=t,$$ $$-\tau $$f(t+2n\tau)=f(t)$$

The period of the function is $$T=2\tau$$

The fourier series of this function is

$$f(t)=\frac{T}{π}[sin(wt)-\frac{sin(2wt)}{2} +\frac{sin(3wt)}{3}-...]$$ What if we let the period $$T$$ to tend to $$\infty$$?

Then, $$w\rightarrow0$$, because w$$=2π\nu=\frac{2π}{T}$$.

Then obviously $$f(t)=t$$ ,for all $$t\in \mathbb R$$

The series becomes like this

$$f(t)=\frac{Tt}{π}[\frac{sin(wt)}{t}-\frac{sin(2wt)}{2t}+\frac{sin(3wt)}{3t}-...]$$, obviously $$t≠0$$ And as we intended, we let $$T\rightarrow \infty$$ and so $$w\rightarrow 0$$

So, $$t=f(t)=\lim\limits_{w \to 0} 2t[\frac{sin(wt)}{wt}-\frac{sin(2wt)}{2wt}+\frac{sin(3wt)}{3wt}-...]=2t[1-1+1-1+1-1+1......]$$ Or, $$t=2t[1-1+1-1+1-1+1....]$$, for all $$t≠0$$

[Disclaimer: We let $$w$$ tend to zero many many faster than the sequence goes to $$\infty$$. In mathematical terms $$\lim\limits_{w\to 0} \frac{sin(nwt)}{nwt}=1$$ for all $$n$$. Means, $$w$$ is such that $$wn\rightarrow0$$ for all $$n$$, as big as possible. Obviously it is our freedom to choose such $$w$$.]

Comparing both sides we get $$\frac{1}{2}=[1-1+1-1+1-1+....]$$ Thus, it's proved.