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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 $n\in \mathbb{N}.$ 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?

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    $\begingroup$ Your argument looks like handwaving, rather than an actual proof. $\endgroup$ – Gabriel Romon Mar 14 '17 at 20:55
  • $\begingroup$ It's not that you're making a mistake, it's that you're not really proving anything. $\endgroup$ – Fimpellizieri Mar 14 '17 at 20:58
  • $\begingroup$ @LeGrandDODOM perhaps I mispoke...it's not that I consider what I've written to be a formal proof, but is the idea right? Or, if I can prove that $\phi$ is just a reordering of the natural numbers then I can easily prove the sums are equal? $\endgroup$ – mizichael Mar 14 '17 at 21:00
  • $\begingroup$ The conclusion is not true if the series is conditionally convergent, so it most definitely does not follow simply from $\phi$ being a reordering. $\endgroup$ – Fimpellizieri Mar 14 '17 at 21:01
  • $\begingroup$ @Fimpellizieri I'm not sure I know how to prove this, then. Absolute convergence of a double sum implies you can switch the order of summation (I forget which theorem gives this). Does the proof then follow directly from that? $\endgroup$ – mizichael Mar 14 '17 at 21:04
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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 $\phi$.

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  • $\begingroup$ I'm always a bit unsure how to work with partial sums as they relate to convergence but I think you've cleared it up a bit, thank you for your help! $\endgroup$ – mizichael Mar 14 '17 at 21:26
  • $\begingroup$ I completely understand! When I first took analysis, partial sums didn't really 'click' with me until I started doing more proofs with them. Let me know if you would like to discuss the second half of this question at all! $\endgroup$ – xk3 Mar 14 '17 at 21:31
  • $\begingroup$ Will do!! Thanks again! $\endgroup$ – mizichael Mar 14 '17 at 21:43
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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 $\sum \vert a_n\vert$. Obviously, $\sum \vert a_n\vert=\sum a_n^+ + \sum a_n^-$. Now, if $\sum \vert a_n\vert$ converges, then none of the two series $\sum a_n^+ $ and $ \sum a_n^-$ may diverge, otherwise $\sum \vert a_n\vert$ 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 $\sum \vert a_n\vert$ 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 $N\to\infty$ if the sum does not converge absolutely. Do notice that $2\mathbb{N}$ has the same cardinality as $\mathbb{N}$, and yet a conditionally convergent series may sum to different valuew when summed over $\mathbb{N}$ and $2\mathbb{N}$; in other words, $\lim\sum^N a_n$ and $\lim\sum^{2N} a_n$ 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.

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