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Before Someone marks this duplicate, this is from an answer just given in this site for another question. But I'm unable to follow this step, how choosing $q<2^n$ forces summation to go on from $k \geq n+1$?

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  • $\begingroup$ what was the original sum? $\endgroup$ – Alain Apr 5 '17 at 8:07
  • $\begingroup$ It is the sum which is subtracted from p/q. It was regarding proving that the sum is irrational. $\endgroup$ – Daniel Evans Apr 5 '17 at 8:09
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    $\begingroup$ I would assume a typo, $\infty$ to be replaced by $n$. Otherwise the LHS is simply $0$. $\endgroup$ – Yves Daoust Apr 5 '17 at 8:22
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    $\begingroup$ In that solution are a few 'clerical' problems that will probably confuse you. First, all the time the solution has a sum that ends in $\infty$, it actually ends in $n$. This is not only true in the place you have marked, but also the next 2 times it comes up. Second, near the end a variable $b$ turns up that should be a $q$. $\endgroup$ – Ingix Apr 5 '17 at 8:42
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    $\begingroup$ I'm talking about the solution, not the problem. Just replace any of the 3 instances where the $\infty$-symbol comes up in the solution with $n$, and the one instance where $b$ comes up with $q$, then you get something that's correct. I'm not saying that the solution does not contain infinite sums, the sum $\sum_{k \ge n+1}\frac1{2^{k^2}}$ from above certainly is infinite. $\endgroup$ – Ingix Apr 5 '17 at 9:00

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