Why can you chose how to align infinitely long equations when adding them?

Question

I saw in a video this proof:

Take this equation:

$$f=1+\frac12+\frac14+\cdots$$

and do this:

$$\begin{align} f&=1+1/2+1/4+\cdots\\ -\quad f/2&=\quad\:\:\:1/2+1/4+\cdots\\ \hline f/2&=1+0+0+\cdots \end{align}$$

This bugs me, and here's why.

$$f/2=1/2+1/4+\cdots$$

We took this equation, but shifted it 1 spot over, effectively making it smaller.

Here's what I mean by shifting. Here's the same thing, but without anything shifted.

$$\begin{align} f&=\:\:\:\:1+1/2+1/4+\cdots\\ -\quad f/2&=1/2+1/4+1/8+\cdots\tag{1}\\ \hline f/2&=1/2+1/4+1/8+\cdots \end{align}$$

1: Now this part isn't shifted anymore.

Basically by shifting, I'm talking about how we align the equations. I don't like the shifting part and here's why. If we take the equation $f=1+1+1+\cdots$ and do a similar thing, first without shifting.

$$\begin{align} f&=1+1+1+\cdots\\ -\quad f&=1+1+1+\cdots\\ \hline 0&=0+0+0+\cdots \end{align}$$

Makes sense, now let's shift it.

$$\begin{align} f&=1+1+1+\cdots\\ -\quad f&=\quad\:\:\:1+1+\cdots\\ \hline 0&=1+0+0+\cdots \end{align}$$

This doesn't make any sense. $0$ doesn't equal $1$. If we were to shift it to the right a second time, we would get $0=2$.

So my question is: why is shifting allowed in that first case? Is there something I'm doing wrong in my example? Am I just not understanding infinity correctly? The way I see it, every time you shift an equation over, you lose a digit, making the answer incorrect.

Answer 1

We can only shift and rearrange infinite sums if both of them converge absolutely. Otherwise strange things happen, just like in your example.

Answer 2

I feel like the answers to this question are not appropriate because they address arbitrary rearrangements of series instead of shifting. Here is an answer that is just concerned with shifting. The key lies in two facts:

Fact 1: Let $S_1 = a_1 + a_2 + \cdots$ and $S_2 = b_1 + b_2 + \cdots$ be two convergent series. Then the sum $(a_1 + b_1) + (a_2 + b_2) + \cdots$ converges and is equal to $S_1 + S_2$.

Fact 2: The series $a_1 + a_2 + \cdots$ converges if and only if the series $0 + a_1 + a_2 + \cdots$ converges.

The second fact is really just justifying the "shift" operation for a series. This proves that if you take two convergent series, you can do all of the shifting, adding (or subtracting) that you want and you will still have a valid answer. Nothing about absolute convergence is necessary for this problem.

This question about the nature of $\infty - \infty$ might be somewhat relevant.

Answer 3

You can prove that conditionally convergent series can be rearranged to converge to any sum while any rearrangement of the terms of the absolutely convergent series must converge to the same limit that the original sequence converges to.

Answer 4

For any infinite sum that's convergent, shifting it doesn't change its value and for any two convergent infinite sums, the difference of their sums is equal to the sum of the differences between the terms. That's true for all convergent infinite sums, not just absolutely convergent sums. What's not true about infinite sums that are convergent but not absolutely convergent is that changing their order can't affect the sum.

Once you've proven that the sum $1 + \frac{1}{2} + \frac{1}{4}...$ is convergent, you can prove using the facts I stated earlier that shifting it over then adding the differences in the terms gives you the result of 1 and therefore the sum of the original sequence is 2. The easiest way to prove that the sum of $1 + \frac{1}{2} + \frac{1}{4}...$ is convergent is probably by first proving it equals 2 so once you've proven that, you're done and there's no need to prove again a different way that the sum is 2.

When you define $f = 1 + 1 + 1...$, you're making the wrong assumption that the infinite sum exists and that that number is called $f$. From the wrong assumption that the sum exists, you can derive the contradiction that $f - f = 0$ and $f - f = 1$.