#reimann's theorem on absolutely convergent series

49 messages · Page 1 of 1 (latest)

mighty shard
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Need help on the proof

zealous berryBOT
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steady forum
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You can start doing the positive case ie $(a_n)$ is a positive sequence in that case $\sum_{n=0}^{\infty} a_n=\sum_{n \in \mathbb{N}} a_n$ and if $\sigma$ is a bijection from $\mathbb{N}$ to itself then $\sum_{n \in \mathbb{N}} a_{\sigma(n)}=\sum_{n \in \sigma \left(\mathbb{N}\right)} a_n$

molten grottoBOT
steady forum
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The first one is also true for any summable family ie if $(b_n){n \in \mathbb{N}}$ is summable then $\sum{n=0}^{\infty} b_n=\sum_{n \in \mathbb{N}} b_n$

molten grottoBOT
steady forum
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Summability being equivalent to absolute convergence of the series for families indexed by $\mathbb{N}$ it follows

molten grottoBOT
steady forum
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Hmmm you know what while what I said is true it feels a bit cheap to directly use a property on summable families so I have another proof I think I can propose that uses the same underlying ideas but more rigorously

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I won’t give the whole proof just the ideas

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Like I said start in the case that (a_n) is a positive sequence

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In that case you can consider $S_n=\sum_{k=0}^{n} a_k$ and $\Sigma_n=\sum_{k=0}^{n} a_{\sigma(k)}$

molten grottoBOT
steady forum
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($\sigma$ here is like before a bijection from $\mathbb{N}$ to itself)

molten grottoBOT
steady forum
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$(S_n){n \in \mathbb{N}}$ and $(\Sigma_n){n \in \mathbb{N}}$ are both increasing sequences ($(a_n)$ having positive values)

molten grottoBOT
steady forum
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In that case we have $\underset{n \to \infty}{lim} S_n=Sup{S_n, n \in \mathbb{N}}$

molten grottoBOT
steady forum
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Likewise $\underset{n \to \infty}{lim} \Sigma_n=Sup{\Sigma_n, n \in \mathbb{N}}$

molten grottoBOT
steady forum
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Calling $L$ and $L’$ their respective limits you can prove that $\forall n \in \mathbb{N}, S_n \le L’$ and that $\forall n \in \mathbb{N}, \Sigma_n \le L$

molten grottoBOT
steady forum
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(Hint: $\sum_{k=0}^{n} a_{\sigma(k)}=\sum_{k \in {1,…,n}}a_{\sigma(k)}$)

molten grottoBOT
steady forum
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via a change of variables it’s equal to $\sum_{k \in \sigma({1,…,n})} a_k$

molten grottoBOT
steady forum
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Sorry for yapping so long I hope it isn’t too confusing

steady forum
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Which concludes the positive case

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I’ll stop typing for now if you need help with the general case tell me (think of how you did to prove that absolute convergence => convergence)

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with (a_n+) and (a_n-)

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@mighty shard

mighty shard
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@steady forum

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I need explanation in that squared part

fluid bronze
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sum a_n converges absolutely, hence its remainder converges to zero, pick large enough N such that the remainder is smaller than epsilon / 2

Then apply triangle inequality

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@mighty shard

mighty shard
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I didn't understand that |Sn - s| part

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How that came

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And how that mod ended to summation

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Ignore what I wrote there in that page

fluid bronze
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@mighty shard

mighty shard
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Yes

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And the next step

fluid bronze
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s_n - s is the remainder term of the series

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it can be made arbitrarily small