Return to Question

I posted this question on math.stackexchange.com and so far the only answer posted (also mentioned in the comments under the question) shows that one of my rash initial guesses about the bottom-line answer was wrong.

Which bijections $f:\{1,2,3,\ldots\}\to\{1,2,3,\ldots\}$ have the property that for every sequence $\{a_n\}_{n=1}^\infty$, $$ \lim_{n\to\infty} \sum_{k=1}^n a_n = \lim_{n\to\infty} \sum_{k=1}^n a_{f(n)}, $$ where "$=$" is construed as meaning that if either limit exists then so does the other and in that case then they are equal?

Here's another rash initial guess, different from the one I posted on stackexchange: It's the bijections whose every orbit is finite.

(That there are uncountably many such bijections can be seen as follows: For each odd number $n$, let $f$ either fix $n$ and $n+1$ or interchange them. That bijection never changes the values of sums. It's a countably infinite sequence of binary choices, so it's uncountable.)

I posted this question on math.stackexchange.com and so far the only answer posted (also mentioned in the comments under the question) shows that one of my rash initial guesses about the bottom-line answer was wrong.

Which bijections $f:\{1,2,3,\ldots\}\to\{1,2,3,\ldots\}$ have the property that for every sequence $\{a_n\}_{n=1}^\infty$, $$ \lim_{n\to\infty} \sum_{k=1}^n a_k = \lim_{n\to\infty} \sum_{k=1}^n a_{f(k)}, $$ where "$=$" is construed as meaning that if either limit exists then so does the other and in that case then they are equal?

Here's another rash initial guess, different from the one I posted on stackexchange: It's the bijections whose every orbit is finite.

(That there are uncountably many such bijections can be seen as follows: For each odd number $n$, let $f$ either fix $n$ and $n+1$ or interchange them. That bijection never changes the values of sums. It's a countably infinite sequence of binary choices, so it's uncountable.)

I posted this question on math.stackexchange.com and so far the only answer posted (also mentioned in the comments under the question) shows that one of my rash initial guesses about the bottom-line answer was wrong.

Which bijections $f:\{1,2,3,\ldots\}\to\{1,2,3,\ldots\}$ have the property that for every sequence $\{a_n\}_{n=1}^\infty$, $$ \lim_{n\to\infty} \sum_{k=1}^n a_n = \lim_{n\to\infty} \sum_{k=1}^n a_{f(n)}, $$ where "$=$" is construed as meaning that if either limit exists then so does the other and in that case then they are equal?

Here's another rash initial guess, different from the one I posted on stackexchange: It's the bijections whose every orbit is finite.

(That there are uncountably many such bijections can be seen as follows: For each odd number $n$, let $f$ either fix $n$ and $n+1$ or interchange them. That bijection never changes the values of sums. It's a countably infinite sequence of binary choices, so it's uncountable.)

I posted this question on math.stackexchange.com and so far the only answer posted (also mentioned in the comments under the question) shows that one of my rash initial guesses about the bottom-line answer was wrong.

Which bijections $f:\{1,2,3,\ldots\}\to\{1,2,3,\ldots\}$ have the property that for every sequence $\{a_n\}_{n=1}^\infty$, $$ \lim_{n\to\infty} \sum_{k=1}^n a_n = \lim_{n\to\infty} \sum_{k=1}^n a_{f(n)}, $$ where "$=$" is construed as meaning that if either limit exists then so does the other and in that case then they are equal?

Here's another rash initial guess, different from the one I posted on stackexchange: It's the bijections whose every orbit is finite.

(That there are uncountably many such bijections can be seen as follows: For each odd number $n$, let $f$ either fix $n$ and $n+1$ or interchange them. That bijection never changes the values of sums. It's a countably infinite sequence of binary choices, so it's uncountable.)