Does a recursive definition of a variable make sense at all?

Question

Let $A:=\sum_{n=0}^\infty 2^n$. I was given the following equation:

$$A=\sum_{n=0}^\infty 2^n=\sum_{n=1}^\infty2^{n-1}=\sum_{n=1}^\infty(2^n\cdot\frac{1}{2})=\frac{1}{2}\sum_{n=1}^\infty2^n=\frac{1}{2}(-1+\sum_{n=0}^\infty2^n)=\frac{1}{2}(-1+A)$$

My task is to explain that why this can't be true and to find the mistake in the equation. I think that I found the mistake but I am not quite sure if I found it correctly so maybe someone can look over it.

First, this obviously has to be wrong, because if I solve $A=\frac{1}{2}(-1+A)$ for $A$ I get $A=-1$. But this can't be true, because $\lim_{n\to\infty}(\sum_{k=1}^n2^k)=+\infty$.

I now tried to check every step and the single steps (except the last) seem to be correct. With the last step I thought that it might be a problem that we have replaced the sum with $A$ again, because then we get a recursive definition of $A$. And from my point of view it makes no sense to do this because then we can never assign a value to $A$. So my thought wass that a recursive definition of a variable is completly senseless.

On the other hand I thought that this might be a notation problem my teacher wants to draw my attention to. In our class we are using the symbol $\sum_{n=0}^\infty a_n$ once for the series and once for the limit $\lim_{n\to\infty}\sum_{k=1}^n a_k$. Maybe we can't replace the $\sum_{n=0}^\infty 2^n$ with $A$ because $A$ is the series and $\sum_{n=0}^\infty 2^n$ is the limit and not the series.

So now I am not quite sure why exactly the equation given is wrong, so maybe someone here can explain better.

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This question here is quite similiar, but from my point of view It wouldn't have helped me to solve my problem, because there is not stated that $A\notin\mathbb{R}$ holds and this was the essential point where I started to understand my problem.

Answer 1

The equation $A=\frac{1}{2}(-1+A)$ is correct, but by solving it in that way, you are assuming that $A$ is a real number (which is not). You should solve it in the extended line $\overline{\mathbb{R}}$ where the equation has actually three solutions (see how arithmetic operations work here): $-1$, $+\infty$ and $-\infty$. Since in this case $A>0$ and the series is not indeterminate (i.e. it does converge to an element of $\overline{\mathbb{R}}$), then the only acceptable solution is $A=+\infty$.

Consider this other example where the series is indeterminate, $$A=\sum_{n=0}^\infty (-1)^n=\sum_{n=1}^\infty(-1)^{n-1}=-\sum_{n=1}^\infty(-1)^n=1-A.$$ The equation $A=1-A$ has a real solution $1/2$ and in $\overline{\mathbb{R}}$ we have also $+\infty$ and $-\infty$, but none of them is acceptable because $\sum_{n=0}^\infty (-1)^n$ should be the limit of $\sum_{n=0}^N(-1)^n$ as $N\to \infty$ which does not exist.

Answer 2

In order to try to keep things clear, I'm going to say "the series $\sum_{n=0}^\infty 2^n$" whenever I'm talking about the series itself (that is, the mere sequence of terms that we are interested in summing), and I'm going to say merely "$\sum_{n=0}^\infty 2^n$" whenever I'm talking about the value (that is, the limit) of the series.

Technically, the mistake you made was the very first sentence of your question: "Let $A:=\sum_{n=0}^\infty 2^n$." If we don't know whether or not the series $\sum_{n=0}^\infty 2^n$ converges, then we don't know whether or not the expression "$\sum_{n=0}^\infty 2^n$" is a defined expression. Since we don't know whether or not that expression is a defined expression, we can't define $A$ as being equal to it.

However, we can fix this mistake by adding one more sentence at the beginning of what you wrote:

Suppose that the series $\sum_{n=0}^\infty 2^n$ converges. Let $A:=\sum_{n=0}^\infty 2^n$. Then

$$A=\sum_{n=0}^\infty 2^n=\sum_{n=1}^\infty2^{n-1}=\sum_{n=1}^\infty(2^n\cdot\frac{1}{2})=\frac{1}{2}\sum_{n=1}^\infty2^n=\frac{1}{2}(-1+\sum_{n=0}^\infty2^n)=\frac{1}{2}(-1+A).$$

Solving the equation $A=\frac{1}{2}(-1+A)$ for $A$, we find that $A = -1$.

All of this is completely true! Every step in the above is correct. It's a totally valid, sound proof. However, it is not a proof that the series $\sum_{n=0}^\infty 2^n$ converges to $-1$, because it has that "suppose" at the beginning of it. This is merely a proof that if the series $\sum_{n=0}^\infty 2^n$ converges, then it converges to -1.

As it happens, the series $\sum_{n=0}^\infty 2^n$ actually doesn't converge. And there is no contradiction between the statement "if the series $\sum_{n=0}^\infty 2^n$ converges, then it converges to -1" and the statement "the series $\sum_{n=0}^\infty 2^n$ does not converge," so we don't have a problem here.

None of this really has anything to do with recursive definitions. The equation $A = \frac12 (-1 + A)$ is not a recursive definition because it's not a definition. If we had tried to define $A$ as $\frac12 (-1 + A)$, then that would have been a recursive definition, and that would have been a problem. But we didn't do that; we defined $A$ differently, and then we did some reasoning which led us to the completely normal and valid equation $A = \frac12 (-1 + A)$.