Prove that ∑n=1∞μ(n)10n is irrational

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Avery Maxwell · Accepted Answer

Step 1The central concern about this problem is to determine whether&amp;nbsp;1+μ(n)&amp;nbsp;is periodic. If it is periodic, we can soon conclude that this number is rational. We will prove the irrationality by contradicting this statement.If&amp;nbsp;1+μ(n)&amp;nbsp;is periodic, then within each period T, there should be only a fixed number of occurences of&amp;nbsp;1+μ(n)=1. This indicates that there should be a fixed number of non square-free integers within the period. However, if we were to let&amp;nbsp;pk&amp;nbsp;be k&#039;th &#039;, we can show that for any&amp;nbsp;M&amp;gt;o&amp;nbsp;the following system of congruences{n≡0&amp;nbsp;(mod&amp;nbsp;pk2)n+1≡0&amp;nbsp;(mod&amp;nbsp;pk+12)n+2≡0&amp;nbsp;(mod&amp;nbsp;pk+22)⋮⋮n+M≡0&amp;nbsp;(mod&amp;nbsp;pk+M2)&amp;nbsp;always has solution by the Chinese remainder theorem, which means that it is possible to have arbitrarily large consecutive non square-free sequence of integers, contradicting our assumption. Therefore, the series∑n=1∞μ(n)10nconverges to an irrational number.

Prove that \(\displaystyle{\sum_{{{n}={1}}}^{{\infty}}}{\frac{{\mu{\left({n}\right)}}}{{{10}^{{n}}}}}\) is irrational

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