Let $A$ be the set $\lbrace n\in \mathbb{N} : n=1$ or the only prime factors of $n$ are $2$ or $3 \rbrace $. So, $6\in A$ but $10\notin A$. Let $S=\sum_{n\in A} \frac{1}{n} $.
What can be said about $S$ ? Can $S$ be real number or $\infty$ ?
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Hint: compare $\;S\;$ to $\;\displaystyle\left(\sum_{j=0}^\infty\,\frac{1}{2^j}\right)\,\left(\sum_{k=0}^\infty\,\frac{1}{3^k}\right)\,$.
dxiv
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Does the order matters while summing? – May 10 '19 at 02:43
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@Unknownx Not for series with all positive terms, since they are absolutely convergent. – dxiv May 10 '19 at 04:06
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This is $$\sum_{j,k=0}^\infty\frac1{2^j3^k} =\left(\sum_{j=0}^\infty\frac1{2^j}\right) \left(\sum_{k=0}^\infty\frac1{3^k}\right).$$ It should be possible to evaluate this explicitly.
Angina Seng
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