Ytukyg

Assume that $sum_{n=1}^{infty} c_n$ is absolutely convergent. Let $phi$ : $mathbb{N}$ → $mathbb{N}$ be a bijection. Set $d_n = c_{phi(n)}$ for $nin mathbb{N}.$ Show that $sum_{n=1}^{infty} d_n$ is convergent, and

$sum_{n=1}^{infty} d_n$ = $sum_{n=1}^{infty} c_n.$

Basically, this problem means that, for an absolutely convergent series, it does not matter what order of summation we are taking as we always get the same result.

I completely understand the intuition here, but it seems that the result is trivial...if $phi$ has the same cardinality as $mathbb{N}$, isn't it really just equivalent to $mathbb{N}$ in a different order? Thus, instead of summing $c_1, c_2, c_3, ...$, we're summing, for example, $c_5, c_{11}, c_{81}, ...$, so that $c_5 + c_{11} + c_{81} + ...$ = $d_1 + d_2 + d_3 + ...$? Is this a sufficient proof? Or, am I making some crucial mistake?

Your intuition on the problem is correct, however, the explanation you provide is not a proof. You can prove that the series $sum_{n=1}^{infty} d_n$ is convergent by appealing to its partial sums which are closely tied to the partial sums of $sum_{n=1}^{infty} c_n$.

Fix $epsilon > 0$. Since $sum_{n=1}^{infty} c_n$ is absolutely convergent, its partial sums are Cauchy. That is, there exists an $N in mathbb{N}$ such that for all $n geq m geq N$, $$|c_m| + |c_{m+1}| + dots |c_n| < epsilon.$$ Then, choose $N_1 > N$ so that if $n geq N_1$ then $phi(n) geq N_1$. Note that since $phi : mathbb{N} rightarrow mathbb{N}$ is a bijection, it is also an injection so there are only a finite number of indices $n$ so that $phi(n) leq N_1$. Then if $k geq ell geq N_1$, we have that $$|c_{phi(ell)}| + |c_{phi(ell+1)}| + dots |c_{phi(k)}| < epsilon$$ i.e., $$|d_{ell}| + |d_{ell+1}| + dots + |d_{k}| < epsilon.$$ Thus the partial sums of $sum_{n=1}^{infty} d_n$ are Cauchy so the sum must converge. To prove that the sums are indeed equal, you should appeal to the surjectivity of $phi$.

You're talking about the Riemann rearrangement theorem: https://en.wikipedia.org/wiki/Riemann_series_theorem

The key point of the Riemann rearrangement theorem was not explained in this wikipedia page, unfortunately. Here it is:

Wikipedia page says "Recall that a conditionally convergent series of real terms has both infinitely many negative terms and infinitely many positive terms.", but without explaining why. So consider the series $sum a_n$. Let $left{ a_n^+right}$ be the sequence of all the positive $a_n$s and let $left{ a_n^-right}$ be the sequence of absolute values of all the negative $a_n$s. Do notice that both $a_n^+$ and $a_n^-$ are positive. Obviously $sum a_n=sum a_n^+ - sum a_n^-$. Now consider the series $sum vert a_nvert$. Obviously, $sum vert a_nvert=sum a_n^+ + sum a_n^-$. Now, if $sum vert a_nvert$ converges, then none of the two series $sum a_n^+ $ and $ sum a_n^-$ may diverge, otherwise $sum vert a_nvert$ would diverge as well as a sum of two positive infinitely large numbers. So immediately one finds that $sum a_n$ converges as well as a difference of two finite numbers. Now, if we drop the condition that $sum vert a_nvert$ converges we're in big trouble, because $sum a_n^+ $ or $ sum a_n^-$ may diverge, because at least one of the series $sum a_n^+ $ and $ sum a_n^-$ is forced to diverge: otherwise the original series would converge absolutely. Why am I mentioning all this? Your proof does not require absolute convergence at all, you do not mention absolute convergence anywhere in your proof. You assume that $phi$ is a permutation and then proceed without mentioning absolute convergence.

The main problem with conditionally convergent series is that if one adds infinitely many infinitely small terms, the sum will change. So, summing over first $N$ numbers and summing over first $2N$ numbers need not yield the same result as $Ntoinfty$ if the sum does not converge absolutely. Do notice that $2mathbb{N}$ has the same cardinality as $mathbb{N}$, and yet a conditionally convergent series may sum to different valuew when summed over $mathbb{N}$ and $2mathbb{N}$; in other words, $limsum^N a_n$ and $limsum^{2N} a_n$ may be different. So one cannot infer much from the fact that $phi$ is a bijection. All the information comes from the fact that the series converges absolutely, not from the cardinality.

I hope this clarifies the situation a bit.