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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?

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    $\begingroup$ 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. $\endgroup$
    – Mufasa
    Oct 29, 2013 at 22:38

3 Answers 3

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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.

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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.

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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 <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.

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