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.