prove that , $2+2+2+2+2+ \cdots= 1+1+1+1+1+\cdots$

Question

how can we prove that ??

i think they are equal

but a friend say that they are not equal

my argument is

$$1+1+1+1+1 + \cdots = \infty$$

$$2+2+2+2+2+\cdots = (1+1) + (1+1) + \cdots = (1+1+1+1+1 + \cdots ) + (1+1+1+1+1 + \cdots ) = \infty + \infty = \infty$$

so they are equal

is this true or not ?

what do you think of my argument ?

and is this equality valid ?

thanx :)

Answer 1

Your equation does not represent finite sums; you are claiming that two infinite series which diverge (blow up as $n \to \infty$) are equal:

$$\sum_{n = 1}^\infty 2\;=\; 2\sum_{n = 1}^\infty 1\longrightarrow +\infty \quad \overset{?} = \quad \sum_{n = 1}^\infty 1 \longrightarrow +\infty$$

Both series are divergent; neither of the sums is equal to $\infty$. They both increase infinitely, which is not to say they equal infinity. Nor are the sums equivalent or "equipotent"; sums, whether finite or infinite, are not sets, so it makes no sense to describe these sums as equal to "the same infinity" or as "equivalent to the same infinity".

Nor to you have asymptotic equivalence: if the partial sum of $n$ terms on your left hand side is given by $f(n) = 2n,\,$ and the partial sum of $n$ terms on your right hand side is given by $g(n) = n$, then $$\lim_{n\to \infty} \frac {f(n)}{g(n)} = \lim_{n \to \infty} \frac {2n}{n} = 2 \neq 1. \;\;\textrm{So}\;\;f(n) \not \asymp g(n).$$

In short, your attempt to prove or disprove your equation is problematic and certainly misguided. It's "not as easy as $1 + 1 = 2$," so to speak.

Answer 2

The usual properties of sum are valid as far as we are dealing with finite sums. A series is not a sum, indeed, they are very different things. Your argument is invalid, I'm sorry to say. Both series are divergent but this does not means that they are the same, this means that their sum is not a finite quantity.

Answer 3

There are many types of infinity, you can't just say that $\infty = \infty$.

For example take $\lim_{x\to \infty} x² = \infty$ and $\lim_{x\to \infty} x = \infty$. We also have $\lim_{x\to \infty} \frac{x}{x²} = \frac{1}{x} = 0$, which just translate the fact that the square function is increasing way faster than the identity function: it doesn't even mean that one infinity is "greater" than the other. You can't use the $=$ operator when dealing with infinity.

You can though say that two kind of infinity are equivalent: for example $\mathbb{N}$ (which is 'infinite') is isomorphic with $\mathbb{Z}$ since there is a bijection between the both of them (for example $f(n) = \lfloor \frac{n+1}{2}*(-1)^n \rfloor$). But it isn't with $\mathbb{R}$ since the functions that go from $\mathbb{R}$ to $\mathbb{N}$ are only surjective and there aren't any surjective function (so no bijection) that go from $\mathbb{N}$ to $\mathbb{R}$. In that sense you could say $\mathbb{R}$ is 'greater' than $\mathbb{N}$.

So the thing is, you gotta be very careful when you want compare infinite quantities. In your case it may not really make sense.

Answer 4

Equality and operations on the divergent series are not allowed only with the meaning of limit of partial sum so for example the equality of the two divergent series $$\sum_{n=1}^\infty\frac{1}{n}=\sum_{n=1}^\infty\log\left(1+\frac{1}{n}\right)$$ and the subtraction $$\sum_{n=1}^\infty\frac{1}{n}-\sum_{n=1}^\infty\log\left(1+\frac{1}{n}\right)$$ are not accepted but we accept $$\lim_{n\to\infty}\left(\sum_{k=1}^n\frac{1}{k}-\sum_{k=1}^n\log\left(1+\frac{1}{k}\right)\right)=\sum_{n=1}^\infty\frac{1}{n}-\log\left(1+\frac{1}{n}\right)$$ which is a convergent series with positive sum.

In the given example: $$1+1+\cdots=2+2+\cdots$$ if the difference is carried out (if we accept the operations) we find a ridiculous result $$0=2+2+\cdots-(1+1+\cdots)=2(1+1+\cdots)-(1+1+\cdots)=(1+1+\cdots)=\infty$$ but all we can do is $$\lim_{n\to\infty}\left(\sum_{k=1}^n2-\sum_{k=1}^n1\right)=\lim_{n\to\infty}\sum_{k=1}^n1=\infty$$

Answer 5

This is has been talked about on math.Stackexchange: When does (Riemann) regularization work?

I have heard of $1 + 2 + 3 + 4 + \dots = -\frac{1}{12}$ this is by ``zeta regularization".

The same method gives $1 + 1 + 1 + 1 + \dots = -\frac{1}{2}$.

Here's an advanced discussion on mathoverflow: Understanding Zeta Function Regularization

We'd like to find the limit as $s \to 0$:

\[\zeta(s) = 1 + \frac{1}{2^s} + \frac{1}{3^s} + \frac{1}{4^s} + \dots = \]

you can use a process called analytic continuation to extend this function to $0$ and to negative numbers.

Then you can ask is analytic continuation a bunch of garbage? It is one consistent way of assigning a number to this sum, using complex analysis... Honestly, I'm not totally convinced this is the best way, but here's a whole Master's thesis on it.

Answer 6

There is a sense in which the given series (i.e. infinite sums) are equal, but it's really just a matter of semantics. For an infinite sequence of positive real numbers $a_{1}, a_{2}, a_{3}\ldots$, we write $a_{1} + a_{2} + a_{3} + \cdots = \infty$ to indicate that the sequence $a_{1}, a_{1} + a_{2}, a_{1} + a_{2} + a_{3}, \ldots$ of finite partial sums grows without bound; i.e. for all $N$, there exists an $n$ such that $a_{1} + a_{2} + a_{3} + \cdots a_{n} > N$. So you could declare that what you mean by $1+1+\cdots = 2+2+\cdots$ is that they both grown without bound (or more technically 'diverge to infinity'). But declaring that they're equal as you did, because of willy-nilly manipulations, is invalid, as was explaind by @Sami Ben Romdhane.