mjhjmtu

I have a short and bitter-sweet riddle for the community this evening. As always, I hope you enjoy it; good luck!

I want to play a game of tag; however, there are some very important rules:

• We both start at the same point.
  
  
  • You cannot tag until the third turn or greater.
  
  


• Each turn you take one step.


• Each turn I double the number of steps I take.
  
  
  • Our stride is exactly the same.
  
  


• We travel in a straight line.


• We both take one initial step.


• Our game is played on an infinitely flat plane


• All steps are forwards, never backwards.
  
  
  • Assume we are infinitely traveling in a straight line.
  
  

Can you tag me?

Please explain your answer.

Hint:

Your step is on top, mine on bottom. Please note the math tag.

$frac11, frac12, frac14, frac18, frac116, frac132...$

So, if we both go

infinitely many times

then your position will be

$x = 1 + 2 + 4 + 8 + 16 + dots$
$x = 1 + 2(1 + 2 + 4 + 8 + dots)$
$x = 1+2x$
$x = -frac12$

and my position will be

$1+1+1+1+1+dots= -frac12$, shown by Dirichlet regularization

hence

I tag you.

The answer is

Yes.

We can do this because:

Geometry. If we were on a plane, you would escape, but if we were on a non-Euclidean surface (like a torus or a sphere) you would come up to me from behind and I could tag you. Since earth is a sphere, this seems to be a safe assumption. This also justifies the knowledge tag.

The answer is:

Yes. Either A tags B before the game starts,

or

A's steps are twice as long as B's, and so tag occurs at the end of round 1.

The answer is

Yes

Why?

I have really long arms

It's not as sensational as the other answers, but there's a genuine, mathematically sound argument to be made that the answer is simply

No.

The distance $d_you$ you've traveled after $n$ turns is $d_you = n^2$, and the distance $d_me$ I've traveled is $d_me = n$.

Using big-O notation, $O(n^2)$ grows faster than $O(n)$, meaning I will never catch you.

One important thing to note that the accepted answer fails to account for:

While $sum_n=1^infty n^2$ and $sum_n=1^infty n$ are both equal to $-frac12$ based on certain mathematical arguments, the naive approaches that argue that they both grow without bound are still valid in other contexts - and I would argue those contexts are the ones that apply to our puzzle here. See the Wikipedia articles (https://en.wikipedia.org/wiki/1_%2B_2_%2B_4_%2B_8_%2B_%E2%8B%AF and https://en.wikipedia.org/wiki/1_%2B_1_%2B_1_%2B_1_%2B_%E2%8B%AF), which say: "Therefore, any totally regular summation method [for $sum_n=1^infty n^2$] gives a sum of infinity," and, "In the context of the extended real number line, $sum_n=1^infty n = infty$, since its sequence of partial sums increases monotonically without bound."

The reason this is the more appropriate way to approach the puzzle is

At any given moment in our game, we'll have only played for a finite amount of time. The game will continue without end forever, but the amount of time we'll have played at any moment is arbitrarily large, which is not the same thing as infinite. This puzzle and this Math.SE question I asked a few years ago explore the distinction and show some of its consequences.

Kinda lame, but

Yes, if you have low stamina and takes some breaks?