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In Elementary Real Analysis of Thomson and Bruckner p85 the proof of $\sum\limits_{i\in\mathbb{Z}}2^{-|i|}=3$ is given:enter image description here

I didn't understand why:$$\sum_{j\in J}2^{-|j|}<2(2^{-N})$$ Could you please help me?

Note: $J$ is a set so that $\forall j\in J,\,j>N$, but I'm personaly convinced that it should be $\forall j\in J,\,|j|>N$ instead especially that otherwise we won't get $I_0\subset J$. Am I right?

Note 2: The definition refered to in the begining of the proof is (in the same page):

enter image description here

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2 Answers 2

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If $\;|j|>N\;\;\;\forall\,j\in J\; $, then

$$2^{|j|}>2^N\;\;\forall\,j\in J\implies 2^{-|j|}<2^{-N}\implies$$

$$\sum_{j\in J}2^{-|j|}=\sum_{j\in J,\,j<0}2^{-|j|}+\sum_{j\in J,\,j\ge 0}2^{-|j|}\le2^{-N}\cdot 2\sum_{k=1}^\infty 2^{-k}=2\cdot2^{-N}$$

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  • $\begingroup$ Thank you very much for your anwser! Unfrontunatly, I didn't got this: $$\sum_{j\in J,\,j<0}2^{-|j|}+\sum_{j\in J,\,j\ge 0}2^{-|j|}\le2^{-N}\cdot 2\sum_{k=1}^\infty 2^{-k}$$. Could you please explain? $\endgroup$
    – Scientifica
    Apr 2, 2015 at 17:52
  • $\begingroup$ @Scientifica Every power in both sums is less than $\;2^{-N};$ , as noted in the first part (or even just the first power $\;2^{-|j|}\;$ for the first $\;j\;$ in the finite set $\;J\;$ , say) and then we can take the sum of all the powers twice $\endgroup$
    – Timbuc
    Apr 2, 2015 at 17:55
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    $\begingroup$ Thank you very much! I got it! $\endgroup$
    – Scientifica
    Apr 2, 2015 at 18:08
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If $J\subset\Bbb Z$ is finite and $j > N$ for all $j\in J$, then there is a $k \in \Bbb N$ such that $J\subset \{N+1,N+2,\ldots, N+k\}$. So

$$\sum_{j\in J} 2^{-|j|} \le \sum_{j = N+1}^{N+k} 2^{-|j|} = \sum_{j = 1}^k 2^{-(N+j)} = 2^{-N-1}\sum_{j = 1}^k 2^{-(j-1)} = 2^{-N}(1 - 2^{-k}) < 2\cdot 2^{-N}$$

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  • $\begingroup$ Thank you very much for your answer! I understood! $\endgroup$
    – Scientifica
    Apr 2, 2015 at 17:50

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